PHYS 202 - GENERAL PHYSICS II -
Kinematics in One Dimension
Question Bank - Set 2
Liberty University
Question 1
Question
A car starts from rest and accelerates at a rate of 2.5 m/s2for 8 seconds. What
is the final velocity of the car?
Solution
Step 1: Identify the knowns and unknowns. The initial velocity uis 0 m/s since
the car starts from rest. The acceleration ais 2.5 m/s2. The time tis 8 seconds.
We need to find the final velocity v.
Step 2: Use the kinematic equation to find the final velocity:
v=u+at
v= 0 m/s + (2.5 m/s2)(8 s)
v= 0 m/s + 20 m/s
v= 20 m/s
Step 3: State the final answer. The final velocity of the car is 20 m/s.
Question 2
Question
A car accelerates uniformly from rest, reaching a speed of 25 m/s in 10 seconds.
What is the acceleration of the car?
Solution
Step 1: Identify the knowns and unknowns.
The initial velocity (v0) is 0 m/s, the final velocity (v) is 25 m/s, and the time
(t) is 10 seconds. We need to find the acceleration (a).
Step 2: Choose a kinematic equation to use.
We can use the kinematic equation:
v=v0+at
Step 3: Plug in the known values.
Substitute v= 25 m/s, v0= 0 m/s, and t= 10 s into the kinematic equation:
25 = 0 + a(10)
Step 4: Solve for acceleration.
Solving for a, we get:
a=25
10
Step 5: Calculate the acceleration.
a= 2.5 m/s2
Step 6: Check the units.
The units for acceleration are in meters per second squared, which is the correct
unit for acceleration.
Question 3
Question
A car accelerates uniformly from rest to a speed of 25 m/s over a distance of
200 m. What is the acceleration of the car?
Solution
Step 1: Identify the given values: The initial velocity (vi) of the car is 0 m/s.
The final velocity (vf) of the car is 25 m/s. The displacement (d) of the car is
200 m.
Step 2: Use the kinematic equation relating initial velocity, final velocity,
acceleration, and displacement:
v2
f=v2
i+ 2ad
Step 3: Substitute the known values into the equation:
(25 m/s)2= (0 m/s)2+ 2a×200 m
2
Step 4: Simplify the equation:
625 = 400a
Step 5: Solve for the acceleration a:
a=625
400 = 1.5625 m/s2
Therefore, the acceleration of the car is 1.5625 m/s2.
Question 4
Question
A car accelerates from rest at a constant rate of 3 m/s2for 10 seconds. After this
time, the car maintains a constant speed for 20 seconds, and then decelerates
at a rate of 2 m/s2until it comes to a stop. What is the total distance traveled
by the car during this entire motion?
Solution
Step 1: Find the distance traveled during acceleration phase.
Time taken during acceleration phase, t1= 10 s
Acceleration, a= 3 m/s2
Initial velocity, u= 0 m/s
Using the kinematic equation s=ut +1
2at2, we find the distance traveled
during acceleration as:
s1= 0 + 1
2·3·(10)2= 150 m
Step 2: Find the distance traveled during constant speed phase.
Time taken during constant speed phase, t2= 20 s
Speed, v= 3 m/s
Distance traveled during constant speed phase is given by:
s2=v·t2= 3 ·20 = 60 m
Step 3: Find the distance traveled during deceleration phase.
Time taken during deceleration phase, t3= 10 s
Deceleration, a= 2 m/s2
3
Speed before deceleration, v= 3 m/s
Using the kinematic equation v2=u2+ 2as, where v= 0 as the car stops
at the end:
0=32+ 2 ·(−2) ·s
Solving for s, we get:
s= 2.25 m
Step 4: Calculate the total distance traveled by the car.
Total distance = s1+s2+s3= 150 + 60 + 2.25 = 212.25 m
Therefore, the total distance traveled by the car during this entire motion is
212.25 meters.
Question 5
Question
A particle moves along a straight line such that its position is given by x(t) =
5t3−2t2+ 3, where xis in meters and tis in seconds. Determine the particle’s
velocity and acceleration as functions of time.
Solution
Step 1: To find the particle’s velocity, we need to take the derivative of the
position function with respect to time (t).
Velocity v(t) = dx
dt
Step 2: Calculate the derivative of x(t):
v(t) = d(5t3−2t2+ 3)
dt
Step 3: Differentiate each term separately:
v(t) = d(5t3)
dt −d(2t2)
dt +d(3)
dt
Step 4: Find the derivatives of each term:
v(t) = 15t2−4t
Step 5: Therefore, the particle’s velocity as a function of time is v(t) =
15t2−4tm/s.
4
Step 6: To find the particle’s acceleration, we need to take the derivative of
the velocity function with respect to time.
Acceleration a(t) = dv
dt
Step 7: Calculate the derivative of v(t):
a(t) = d(15t2−4t)
dt
Step 8: Differentiate each term separately:
a(t) = d(15t2)
dt −d(4t)
dt
Step 9: Find the derivatives of each term:
a(t) = 30t−4
Step 10: Therefore, the particle’s acceleration as a function of time is a(t) =
30t−4 m/s2.
Question 6
Question
A car accelerates from rest at a constant rate of 2 m/s2. How fast is the car
going after 5 seconds?
Solution
Step 1: First, we need to determine the final velocity of the car using the
kinematic equation:
v=u+at
where v= final velocity, u= initial velocity, a= acceleration, and t= time.
Given that u= 0 (the car starts from rest), a= 2 m/s2, and t= 5 s, we can
plug these values into the equation:
v= 0 + 2 ×5
v= 10 m/s
Therefore, the car is going 10 m/s after 5 seconds.
5
Question 7
Question
A car accelerates uniformly from rest at 2.5 m/s2for 5 seconds, then continues
at a constant velocity for 10 seconds, and finally decelerates uniformly to come
to a stop in 4 seconds. Determine the total distance traveled by the car during
this time.
Solution
Step 1: Calculate the distance traveled during acceleration phase. Given that
initial velocity (vi) is 0 m/s, acceleration (a) is 2.5 m/s2, and time (t) is 5
seconds. We can use the kinematic equation: d=vit+1
2at2. Substitute the
values and calculate:
d= (0 ×5) + 1
2×2.5×52
d= 0 + 1
2×2.5×25
d= 0 + 31.25
d= 31.25 m
Step 2: Calculate the distance traveled during constant velocity phase. Since
the car is traveling at a constant velocity, the distance traveled is simply the
product of velocity and time. Given that the velocity (v) during this phase is
2.5 m/s and time (t) is 10 seconds:
d=v×t
d= 2.5×10
d= 25 m
Step 3: Calculate the distance traveled during deceleration phase. The car
comes to a stop during this phase, so the distance traveled can be calculated
using the same formula we used for acceleration. The final velocity (vf) is 0
m/s, acceleration (a) is -2.5 m/s2, and time (t) is 4 seconds:
d=vit+1
2at2
d= 0 ×4 + 1
2×(−2.5) ×42
d= 0 + 1
2×(−2.5) ×16
d= 0 −20
d=−20 m
6
Step 4: Calculate the total distance traveled by the car. The total distance
traveled is the sum of the distances calculated in each phase. Total distance =
31.25 m + 25 m - 20 m Total distance = 36.25 m
Therefore, the total distance traveled by the car during this time is 36.25
meters.
Question 8
Question
A car is traveling along a straight road. The car accelerates from rest at a
constant rate of 2 m/s2for 10 seconds, then maintains a constant speed for 20
seconds, and finally decelerates at a rate of −3 m/s2until it comes to a stop.
Determine the total distance the car travels during this time interval.
Solution
Step 1: Calculate the distance traveled during the acceleration phase. The
distance traveled during constant acceleration can be calculated using the equa-
tion:
d=1
2at2
where ais the acceleration and tis the time interval.
Given a= 2 m/s2and t= 10 s:
d=1
2×2 m/s2×(10 s)2
d= 100 m
Step 2: Calculate the distance traveled during the constant speed phase.
Since the speed is constant, the distance traveled is given by:
distance = speed ×time
Given speed = 20 m/s and time = 20 s:
distance = 20 m/s ×20 s
distance = 400 m
Step 3: Calculate the distance traveled during the deceleration phase. The
distance traveled during deceleration can be calculated using the equation:
d=vit+1
2at2
where viis the initial velocity, tis the time interval, and ais the acceleration.
7
Given vi= 20 m/s, a=−3 m/s2, and t= 10 s:
d= 20 m/s ×10 s + 1
2× −3 m/s2×(10 s)2
d= 200 m −150 m
d= 50 m
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of the distances during each phase:
Total distance = 100 m + 400 m + 50 m
Total distance = 550 m
Therefore, the total distance the car travels during this time interval is 550
meters.
Question 9
Question
A car is traveling along a straight road. Initially, the car is moving with a
velocity of 15 m/s, and then it accelerates at a constant rate of 2 m/s2for 8
seconds. After this time, the car brakes and decelerates at a constant rate of
3 m/s2until it comes to a stop. Determine the total distance traveled by the
car during this entire motion.
Solution
Step 1: Find the distance covered during acceleration phase. Using the equation
of motion: v=u+at, where vis the final velocity, uis the initial velocity, a
is the acceleration, and tis the time, the final velocity after 8 seconds can be
found as: v= 15 + 2(8) = 31 m/s.
The distance covered during acceleration phase can be calculated using the
formula: s=ut +1
2at2, where sis the distance, uis the initial velocity, ais the
acceleration, and tis the time, s= 15(8) + 1
2·2·(8)2= 120 + 64 = 184 m.
Step 2: Find the distance covered during deceleration phase. The final
velocity after decelerating to stop can be found using the equation of motion:
v2=u2+ 2as, where vis the final velocity (0 in this case), uis the initial
velocity (31 m/s), ais the acceleration (3 m/s2), and sis the distance, 0 =
(31)2+ 2 ·(−3) ·s. Solving for sgives: s=(31)2
2·3=961
6= 160.17 m.
Step 3: Find the total distance traveled. The total distance traveled by
the car is the sum of the distances covered during acceleration and deceleration
phases: Total distance = 184 m + 160.17 m = 344.17 m.
Therefore, the total distance traveled by the car during its entire motion is
344.17 meters.
8
Question 10
Question
A ball is thrown vertically upward with an initial velocity of 30 m/s. How long
does it take for the ball to reach its maximum height? What is the maximum
height the ball reaches? (Assume the acceleration due to gravity is 9.81 m/s2)
Solution
Step 1: Identify the known variables and the equation to be used.
Initial velocity (vi) = 30 m/s
Acceleration due to gravity (g) = 9.81 m/s2
Final velocity at the maximum height (vf) = 0 m/s (ball momentarily
stops at the maximum height)
We will use the kinematic equation for motion with constant acceleration:
vf=vi+at where vf= 0.
Step 2: Find the time it takes for the ball to reach the maximum height.
Using the kinematic equation vf=vi+at with vf= 0:
0 = 30 + (−9.81)t
9.81t= 30
t=30
9.81 ≈3.06 s
Step 3: Find the maximum height the ball reaches. We can use the kinematic
equation for displacement: s=vit+1
2at2. Substitute vi= 30 m/s, t= 3.06 s,
and a=−9.81 m/s2into the equation:
s= 30(3.06) + 1
2(−9.81)(3.06)2
s= 91.8−45.2
s≈46.6 m
Therefore, it takes approximately 3.06 seconds for the ball to reach its max-
imum height, and the maximum height the ball reaches is approximately 46.6
meters.
Question 11
Question
A car initially at rest starts moving in a straight line with a constant acceleration
of 3 m/s2. How long will it take for the car to reach a speed of 30 m/s?
9
Solution
Step 1: Let’s denote the initial velocity of the car as v0, the final velocity as vf,
the acceleration as a, and the time taken to reach the final velocity as t. Given:
v0= 0, a= 3 m/s2, and vf= 30 m/s.
Step 2: We can use the kinematic equation:
vf=v0+a·t
Substitute the known values:
30 = 0 + 3t
Step 3: Solve for t:
t=30
3= 10 seconds
Step 4: Therefore, it will take the car 10 seconds to reach a speed of 30 m/s.
Question 12
Question
A car is traveling along a straight road. The car starts from rest and accelerates
uniformly at 2.5 m/s2for 10 seconds. After this time, the car maintains a
constant velocity for another 20 seconds. Finally, the car decelerates uniformly
to a stop in 5 seconds. Calculate the total distance the car travels during this
entire period.
Solution
Step 1: Calculate the distance traveled during acceleration phase. The distance
traveled during acceleration can be calculated using the kinematic equation:
d=1
2·a·t2
Substitute a= 2.5 m/s2and t= 10 s into the equation:
d=1
2·2.5 m/s2·(10 s)2
d=1
2·2.5 m/s2·100 s2
d= 125 m
So, the car travels 125 meters during the acceleration phase.
Step 2: Calculate the distance traveled during constant velocity phase. Dur-
ing the constant velocity phase, the car moves at a constant speed, which can
be calculated by multiplying the constant velocity by the time:
d=v·t
10
Substitute the constant velocity and the time:
d= 25 m/s ·20 s
d= 500 m
So, the car travels 500 meters during the constant speed phase.
Step 3: Calculate the distance traveled during deceleration phase. The dis-
tance traveled during the deceleration phase can be calculated using the kine-
matic equation:
d=1
2·a·t2
Substitute a=−2.5 m/s2(as it’s deceleration) and t= 5 s into the equation:
d=1
2·(−2.5) m/s2·(5 s)2
d=1
2·(−2.5) m/s2·25 s2
d=−31.25 m
So, the car travels 31.25 meters during the deceleration phase.
Step 4: Calculate the total distance traveled by the car. To calculate the
total distance, we sum up the distances traveled during each phase:
T otal distance = 125 m + 500 m + 31.25 m
T otal distance = 656.25 m
Therefore, the car travels a total distance of 656.25 meters.
Question 13
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2for 10
seconds. After this time, the car maintains a constant speed for the next 20
seconds. Finally, the car decelerates at a rate of 1.5 m/s2until it comes to a
stop. Determine the total distance traveled by the car during this entire trip.
Solution
Step 1: Find the distance traveled during acceleration phase.
The distance traveled during acceleration is given by the formula:
d=1
2×acceleration ×time2
11
Substitute the values: acceleration = 2.5 m/s2, time = 10 seconds.
d=1
2×2.5×102
d=1
2×2.5×100
d= 125 m
Therefore, the distance traveled during the acceleration phase is 125 m.
Step 2: Find the distance traveled during constant speed phase.
During constant speed, the distance traveled is given by the formula:
d= speed ×time
We know the car maintained a speed, so the distance traveled during this
phase is:
d= speed ×time
d= speed ×20
Step 3: Find speed during constant speed phase.
Since the car maintained a constant speed during the 20 seconds, the speed
is the final speed at the end of the acceleration phase. The final speed is given
by:
Final speed = initial speed + acceleration ×time
Substitute the values: initial speed = 0 m/s, acceleration = 2.5 m/s2, time
= 10 seconds.
Final speed = 0 + 2.5×10
Final speed = 0 + 25
Final speed = 25 m/s
Therefore, the speed during the constant speed phase is 25 m/s.
Step 4: Calculate the distance traveled during constant speed phase.
d= speed ×20
d= 25 ×20
d= 500 m
Therefore, the distance traveled during the constant speed phase is 500 m.
Step 5: Find the distance traveled during the deceleration phase.
The distance traveled during deceleration is given by the formula:
d=1
2×deceleration ×time2
Substitute the values: deceleration = 1.5 m/s2, time = unknown, final speed
= 0 m/s.
0 = 25 + (−1.5) ×time
12
25 = 1.5×time
time = 25
1.5
time = 16.67 seconds
Therefore, the total time for deceleration is approximately 16.67 seconds.
Now, substitute the time into the distance formula:
d=1
2×1.5×(16.67)2
d=1
2×1.5×278.89
d= 208.67 m
Thus, the distance traveled during the deceleration phase is 208.67 m.
Step 6: Calculate the total distance traveled by the car.
The total distance traveled is the sum of the distances traveled during each
phase:
Total distance = 125 + 500 + 208.67
Total distance = 833.67 m
Therefore, the total distance traveled by the car during this entire trip is
approximately 833.67 meters.
Question 14
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the car continues at a constant velocity for another 20 seconds.
If the total distance traveled by the car during this time is 500 meters, find the
distance traveled during the constant velocity phase.
Solution
Step 1: Find the distance traveled during the acceleration phase. To find the
distance traveled during the acceleration phase, we first need to find the final
velocity of the car after accelerating for 10 seconds. We can use the equation
of motion: v=u+at, where: - vis the final velocity, - uis the initial velocity
(which is 0 m/s since the car starts from rest), - ais the acceleration of the
car (which is 2 m/s2), and - tis the time the car accelerates for (which is 10
seconds).
Substitute the values into the equation to find v:
v= 0 + (2 m/s2)(10 s) = 20 m/s
13
Next, use the formula for distance traveled during constant acceleration mo-
tion: s=ut +1
2at2, where: - sis the distance traveled, - uis the initial velocity,
-tis the time, and - ais the acceleration.
Substitute the values into the formula to find the distance traveled during
the acceleration phase:
sacc = (0 m/s)(10 s) + 1
2(2 m/s2)(10 s)2= 100 m
So, the distance traveled during the acceleration phase is 100 meters.
Step 2: Find the distance traveled during the constant velocity phase. Since
the total distance traveled by the car is 500 meters and the distance traveled
during the acceleration phase is 100 meters, the distance traveled during the
constant velocity phase can be found by subtraction:
sconst = 500 m −100 m = 400 m
Therefore, the distance traveled during the constant velocity phase is 400
meters.
Question 15
Question
A car accelerates from rest at a constant rate of 2.5 m/s2for 10 seconds. After
this time, what is its displacement from the starting point?
Solution
Step 1: Identify the given variables. Given: Initial velocity, u= 0 m/s (as the
car starts from rest); Acceleration, a= 2.5 m/s2; Time, t= 10 s.
Step 2: Use the kinematic equation for displacement. The equation for
displacement in terms of initial velocity, acceleration, and time is:
s=ut +1
2at2
Step 3: Substitute the given values into the equation.
s= (0)(10) + 1
2(2.5)(10)2
Step 4: Calculate the displacement.
s= 0 + 1
2(2.5)(100)
s=1
2×250
s= 125 meters
Step 5: Answer After accelerating at 2.5 m/s2for 10 seconds, the car’s
displacement from the starting point is 125 meters.
14
Question 16
Question
A car accelerates uniformly from rest at 2.0 m/s2. How far will the car have
traveled when it reaches a velocity of 25 m/s?
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, u= 0 m/s (rest) Acceleration, a= 2.0 m/s2Final
velocity, v= 25 m/s Distance traveled, x= ?
Step 2: Determine the time taken to reach the final velocity.
The final velocity can be related to the initial velocity, acceleration, and time
using the kinematic equation v=u+at, where:
v=u+at
25 = 0 + 2t
t=25
2
t= 12.5 seconds
Step 3: Calculate the distance traveled using the equation x=ut +1
2at2.
Substitute the known values into the equation:
x= 0(12.5) + 1
2(2)(12.5)2
x= 0 + 0.5(2)(156.25)
x= 0 + 156.25
x= 156.25 meters
Therefore, the car will have traveled 156.25 meters when it reaches a velocity
of 25 m/s.
Question 17
Question
A car starts from rest at a traffic light and accelerates uniformly at 3.0 m/s2for
8.0 seconds. After this, the car continues at a constant velocity for an additional
10 seconds. Determine the total distance the car travels during this time period.
15
Solution
Step 1: Find the distance traveled during the acceleration phase. The distance
traveled during the acceleration phase can be found using the equation:
d1=1
2at2
1
where: a= 3.0 m/s2(acceleration), t1= 8.0 s (time during acceleration).
Substitute the values and solve for d1:
d1=1
2×3.0 m/s2×(8.0 s)2
d1= 0.5×3.0 m/s2×64.0 s2
d1= 96.0 m
Therefore, the distance traveled during the acceleration phase is 96.0 meters.
Step 2: Find the distance traveled during the constant velocity phase. Dur-
ing the constant velocity phase, the distance traveled is given by:
d2=v×t2
where: vis the constant velocity, t2= 10 s (time during constant velocity).
Since the car continued at a constant velocity after the acceleration phase,
the final velocity at the end of the acceleration phase is the velocity during the
constant velocity phase. We can find the final velocity using:
vf=a×t1
vf= 3.0 m/s2×8.0 s
vf= 24.0 m/s
Substitute the values to find d2:
d2= 24.0 m/s ×10 s
d2= 240.0 m
Therefore, the distance traveled during the constant velocity phase is 240.0
meters.
Step 3: Find the total distance traveled. The total distance traveled is
the sum of the distances traveled during the acceleration and constant velocity
phases:
Total distance = d1+d2
Total distance = 96.0 m + 240.0 m
Total distance = 336.0 m
Hence, the total distance the car travels during this time period is 336.0
meters.
16
Question 18
Question
A car accelerates uniformly from rest to a speed of 30 m/s over a distance of
150 m. What is the magnitude of the car’s acceleration?
Solution
Step 1: Identify the given information.
The final velocity of the car, vf, is 30 m/s. The initial velocity, vi, is 0 m/s
(since the car starts from rest). The displacement of the car, ∆x, is 150 m.
Step 2: Determine the acceleration using the kinematic equation:
v2
f=v2
i+ 2a∆x
Step 3: Plug in the known values and solve for acceleration:
a=v2
f−v2
i
2∆x
a=(30 m/s)2−(0 m/s)2
2(150 m)
a=900 m2/s2
300 m
a= 3 m/s2
Therefore, the magnitude of the car’s acceleration is 3 m/s2.
Question 19
Question
A car accelerates from rest at a constant rate of 2.0 m/s2along a straight road.
After 5.0 seconds, what is the displacement of the car?
Solution
Step 1: First, we need to determine the final velocity of the car after 5.0 seconds
using the kinematic equation:
v=v0+at
where vis the final velocity, v0is the initial velocity, ais the acceleration, and t
is the time. Given that the initial velocity v0= 0 m/s, the acceleration a= 2.0
m/s2, and the time t= 5.0 s, we have:
v= 0 + (2.0 m/s2)(5.0 s) = 10 m/s
17
Step 2: Next, we can find the displacement of the car using the kinematic
equation:
s=v0t+1
2at2
where sis the displacement, v0is the initial velocity, ais the acceleration, and t
is the time. Given that the initial velocity v0= 0 m/s, the acceleration a= 2.0
m/s2, and the time t= 5.0 s, we have:
s= 0(5.0 s) + 1
2(2.0 m/s2)(5.0 s)2= 25 m
Therefore, the displacement of the car after 5.0 seconds is 25 meters.
Question 20
Question
A car is traveling along a straight road at a constant velocity of 25 m/s for 20
seconds. It then accelerates uniformly to a velocity of 40 m/s over the next 10
seconds. Finally, it maintains this velocity for an additional 30 seconds before
coming to a stop with a constant deceleration. Find the total distance the car
traveled during this entire time.
Solution
Step 1: Find the distance traveled during the first phase of motion (constant
velocity). The distance traveled during constant velocity can be found using
the formula:
d=v×t
where dis the distance, vis the velocity, and tis the time. Given v= 25 m/s
and t= 20 seconds:
d= 25 ×20 = 500 meters
Step 2: Find the distance traveled during the second phase of motion (uni-
form acceleration). The distance traveled during uniform acceleration can be
found using the formula:
d=vi×t+1
2×a×t2
where dis the distance, viis the initial velocity, ais the acceleration, and tis
the time. Given vi= 25 m/s, a=40−25
10 = 1.5 m/s2, and t= 10 seconds:
d= 25 ×10 + 1
2×1.5×(10)2= 250 + 75 = 325 meters
Step 3: Find the distance traveled during the third phase of motion (constant
velocity). Similar to Step 1: Given v= 40 m/s and t= 30 seconds:
d= 40 ×30 = 1200 meters
18
Step 4: Find the distance traveled during the fourth phase of motion (de-
celeration). Since the car comes to a stop with constant deceleration, the total
distance traveled during deceleration is equal to the distance traveled in the
third phase (since the car comes to a stop). Therefore, the distance traveled
during deceleration is 1200 meters.
Step 5: Add up the distances from each phase to find the total distance
traveled. The total distance traveled is the sum of the distances from each
phase:
500 + 325 + 1200 + 1200 = 3225 meters
Question 21
Question
A car accelerates from rest at a constant rate of 3 m/s2for 10 seconds. After
this time, the car slows down with a constant acceleration of −2 m/s2until it
comes to a stop. Determine the total distance travelled by the car during this
motion.
Solution
Step 1: Find the distance travelled during the acceleration phase. The distance
travelled during acceleration can be found using the kinematic equation:
d1=vit+1
2a1t2
where - d1is the distance travelled during acceleration, - viis the initial velocity
during acceleration (which is 0 since the car starts from rest), - a1is the accel-
eration during the acceleration phase (which is 3 m/s2), - tis the time during
which the car accelerates (which is 10 seconds).
Plugging in the values, we get:
d1= 0 ×10 + 1
2×3×(10)2
d1= 0 + 1
2×3×100
d1= 0 + 1.5×100
d1= 150 m
Step 2: Find the distance travelled during the deceleration phase. The
distance travelled during deceleration can also be found using the kinematic
equation, however, since the car comes to a stop, the final velocity during de-
celeration is 0. Therefore, the distance travelled during deceleration can be
expressed as:
d2=v2
f−v2
i
2a2
19
where - d2is the distance travelled during deceleration, - vfis the final velocity
during deceleration (which is 0), - viis the initial velocity during deceleration,
-a2is the acceleration during the deceleration phase (which is −2 m/s2).
During the acceleration phase, the final velocity was:
vi=vi+a1t= 0 + 3 ×10 = 30 m/s
Plugging in the values, we get:
d2=(0)2−(30)2
2× −2
d2=0−900
−4
d2=−900
−4
d2= 225 m
Step 3: Find the total distance travelled by the car. The total distance
travelled by the car is the sum of the distances travelled during acceleration
and deceleration:
Total distance = d1+d2= 150 m + 225 m = 375 m
Therefore, the total distance traveled by the car during this motion is 375 m .
Question 22
Question
A car is initially traveling at v0= 20 m/s and accelerates uniformly at a=
3 m/s2for a distance of d= 100 m. What is the final velocity of the car?
Solution
Step 1: Calculate the final velocity using the equation of motion for uniformly
accelerated motion:
Final velocity2= Initial velocity2+ 2 ×acceleration ×distance
Final velocity = qv2
0+ 2ad
Step 2: Substitute the given values into the equation:
Final velocity = q(20 m/s)2+ 2 ×(3 m/s2)×(100 m)
Step 3: Perform the calculation:
Final velocity = √400 + 600
Final velocity = √1000
Final velocity = 31.62 m/s
Therefore, the final velocity of the car is 31.62 m/s.
20
Question 23
Question
A car accelerates from rest at a rate of 3 m/s2for 10 seconds. After this time,
the car maintains a constant velocity for an additional 20 seconds. Finally, the
car decelerates at a rate of 2 m/s2until it comes to a stop. Calculate the total
distance traveled by the car during this entire process.
Solution
Step 1: Find the distance traveled during the acceleration phase. The distance
traveled during acceleration can be found using the equation:
d=1
2at2
where dis the distance, ais the acceleration, and tis the time. Plugging in the
values, we get:
d=1
2×3×(10)2
d=1
2×3×100
d= 150 m
Step 2: Find the distance traveled during the constant velocity phase. Since
the velocity is constant, the distance traveled can be calculated using the equa-
tion:
d=vt
where dis the distance, vis the velocity, and tis the time. Given that the car
maintains a constant velocity for 20 seconds, the distance traveled during this
phase is:
d=v×20
Since the initial velocity is 30 m/s (acquired during the acceleration phase), the
distance traveled during this phase is:
d= 30 ×20
d= 600 m
Step 3: Find the distance traveled during the deceleration phase. The dis-
tance traveled during deceleration can be found using the equation:
d=vit+1
2at2
where dis the distance, viis the initial velocity, ais the acceleration, and t
is the time. Given that the initial velocity is 30 m/s (as the car maintains a
21
constant velocity of 30 m/s for 20 seconds) and the deceleration rate is −2 m/s2,
the distance traveled during this phase is:
d= 30 ×20 + 1
2×(−2) ×(20)2
d= 600 −200
d= 400 m
Step 4: Calculate the total distance traveled. The total distance traveled by
the car is the sum of the distances traveled in each phase:
Total distance = 150 + 600 + 400
Total distance = 1150 m
Therefore, the total distance traveled by the car during this process is 1150
meters.
Question 24
Question
A car starts from rest and accelerates at a constant rate of 2.0 m/s2. How fast
is it going after it has traveled 100 meters? (The car starts at x= 0, with initial
velocity v0= 0, and acceleration a= 2.0 m/s2)
Solution
Step 1: Use the kinematic equation for velocity to find the final velocity of the
car.
v=v0+at
Since v0= 0 and a= 2.0 m/s2, we have
v= 0 + 2.0 m/s2·t
Step 2: Find the time ttaken for the car to travel 100 meters using the
kinematic equation for position.
x=x0+v0t+1
2at2
Since x= 100 m, x0= 0, v0= 0, and a= 2.0 m/s2, we have
100 = 0 + 0 + 1
2·2.0 m/s2·t2
Solving for t:
100 = t2
22
t= 10 s
Step 3: Substitute t= 10 s into the expression for velocity to find the final
velocity v.
v= 2.0 m/s2·10 s
v= 20 m/s
Answer: The car is going 20 m/s after traveling 100 meters.
Question 25
Question
A car accelerates from rest at a constant rate of 2 m/s2for a distance of 50
meters before reaching a constant velocity. How long does it take for the car to
reach this constant velocity?
Solution
Step 1: We can find the time it takes for the car to reach its constant velocity
using the kinematic equation for motion in one dimension:
v=u+at
where: - vis the final velocity, - uis the initial velocity, - ais the acceleration,
and - tis the time taken.
Step 2: Since the car starts from rest, the initial velocity u= 0 m/s.
Step 3: For the accelerating phase, we are given:
a= 2 m/s2
u= 0 m/s
v=?
Step 4: Using the equation v=u+at, we can solve for v:
v= 0 + (2)(t)
v= 2t
Step 5: During the acceleration phase, the car covers a distance of 50 meters.
We know that distance = average velocity ×time.
Step 6: The average velocity during acceleration is given by:
Average velocity = u+v
2
Average velocity = 0+2t
2=t
23
Step 7: Since the car travels 50 meters during the acceleration phase:
50 = t×t
50 = t2
Step 8: Solving for t:
t=√50
t≈7.07 s
Step 9: Therefore, it takes approximately 7.07 seconds for the car to reach
its constant velocity.
Question 26
Question
A car initially traveling at a speed of 20 m/s undergoes constant acceleration
and comes to a stop in 100 meters. What is the magnitude of the acceleration
of the car?
Solution
Step 1: Identify the known quantities. The initial velocity of the car, u, is 20
m/s, the final velocity of the car, v, is 0 m/s, the displacement of the car, s, is
100 meters.
Step 2: Choose the appropriate kinematic equation. We can use the equation
that relates initial velocity, final velocity, acceleration, and displacement without
time:
v2=u2+ 2as
Step 3: Plug in the known values and solve for the acceleration. Substitute
v,u, and sinto the equation:
0 = (20)2+ 2a(100)
0 = 400 + 200a
Solving for a:
a=−400
200 =−2 m/s2
Step 4: State the answer. The magnitude of the acceleration of the car is 2
m/s2.
24
Question 27
Question
A particle moves along a straight line according to the equation x= 4t2−3t+5,
where xis in meters and tis in seconds. Find the acceleration of the particle
when t= 3 s.
Solution
Step 1: Find the velocity of the particle using the derivative of the position
function.
v(t) = dx
dt =d
dt(4t2−3t+ 5)
v(t)=8t−3
Step 2: Find the acceleration of the particle using the derivative of the
velocity function.
a(t) = dv
dt =d
dt(8t−3)
a(t)=8
Step 3: Calculate the acceleration of the particle at t= 3 s.
a(3) = 8 m/s2
Therefore, the acceleration of the particle at t= 3 s is 8 m/s2.
Question 28
Question
A car is initially traveling at a speed of 20 m/s. It then accelerates at a constant
rate of 3 m/s2for 8 seconds. What is the final velocity of the car after this
acceleration period?
Solution
Step 1: Determine the acceleration of the car using the equation a=∆v
∆t, where
ais the acceleration, ∆vis the change in velocity, and ∆tis the time interval.
Given: Initial velocity vi= 20 m/s, acceleration a= 3 m/s2, time interval
∆t= 8 s. Plug in the values: a=vf−vi
∆t3 = vf−20
824 = vf−20 vf= 24 + 20
vf= 44 m/s
Therefore, the final velocity of the car after accelerating at a constant rate
of 3 m/s2for 8 seconds is 44 m/s.
25
Question 29
Question
A car starts from rest and accelerates along a straight road according to the
equation x(t)=2t2+ 3t+ 1, where xis in meters and tis in seconds. Calculate
the acceleration of the car at t= 2 seconds.
Solution
Step 1: To find the acceleration of the car, we need to differentiate the displace-
ment function x(t) with respect to time tto get the velocity function, and then
differentiate the velocity function to get the acceleration function.
Step 2: Let’s start by finding the velocity function v(t) by differentiating
x(t) with respect to t:
v(t) = dx
dt =d
dt(2t2+ 3t+ 1) = 4t+ 3
Step 3: Next, let’s find the acceleration function a(t) by differentiating v(t)
with respect to t:
a(t) = dv
dt =d
dt(4t+ 3) = 4
Step 4: Now, to find the acceleration of the car at t= 2 seconds, substitute
t= 2 into the acceleration function:
a(2) = 4 m/s2
Therefore, the acceleration of the car at t= 2 seconds is 4 m/s2.
Question 30
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 10
seconds. After this time, the car maintains a constant velocity for an additional
20 seconds. Finally, the car decelerates at a rate of 2.0 m/s2until it comes to a
stop. Determine the total distance the car travels during this entire process.
Solution
Step 1: Determine the distance traveled during acceleration phase. Given that
the car starts from rest and accelerates at 3.0 m/s2for 10 seconds, we can use
the equation x=vit+1
2at2, where xis the distance traveled, viis the initial
velocity (0 m/s), ais the acceleration, and tis the time.
x= 0 ·10 + 1
2·3.0·(10)2= 150 m
26
Step 2: Determine the distance traveled during constant velocity phase.
During this phase, the car travels at a constant velocity for 20 seconds. Since
velocity is constant, distance traveled can be calculated by multiplying velocity
by time.
x= 3.0·20 = 60 m
Step 3: Determine the distance traveled during deceleration phase. In this
phase, the car decelerates at 2.0 m/s2until it comes to a stop. We can again
use the equation x=vit+1
2at2, where viis the initial velocity (3.0 m/s), ais
the deceleration (-2.0 m/s2), and tis the time. The time taken to stop can be
calculated as t=vf−vi
a=0−3.0
−2.0= 1.5 seconds. The distance traveled during
deceleration phase is then
x= 3.0·1.5 + 1
2·(−2.0) ·(1.5)2= 4.5−1.125 = 3.375 m
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of the distances traveled during each phase.
Total Distance = 150 + 60 + 3.375 = 213.375 m
Therefore, the total distance the car travels during this entire process is
213.375 meters.
Question 31
Question
A car starts from rest and accelerates at a constant rate of 3.50 m/s2. How long
does it take for the car to reach a speed of 28.0 m/s?
Solution
Step 1: We can use the kinematic equation to solve for the time it takes for the
car to reach a speed of 28.0 m/s. The kinematic equation relating final velocity,
initial velocity, acceleration, and time is given by:
v=u+at
where: v= final velocity (28.0 m/s), u= initial velocity (0 m/s, as the car
starts from rest), a= acceleration (3.50 m/s2), t= time.
Step 2: Substituting the given values into the kinematic equation, we have:
28.0 m/s = 0 m/s + 3.50 m/s2·t
Step 3: Solve for tby isolating it on one side of the equation:
t=28.0 m/s
3.50 m/s2
27
Step 4: Calculate the time it takes for the car to reach a speed of 28.0 m/s:
t=28.0
3.50 = 8.00 s
Answer: The car will take 8.00 seconds to reach a speed of 28.0 m/s.
Question 32
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2along a straight
road. How long does it take for the car to reach a speed of 20 m/s?
Solution
Step 1: Identify the given quantities and the unknown. Let: - The initial velocity
of the car be vi= 0 m/s (starting from rest). - The acceleration of the car be
a= 3.0 m/s2. - The final velocity of the car be vf= 20 m/s. - The time taken
for the car to reach the final velocity be t(unknown).
Step 2: Use the kinematic equation relating initial velocity, final velocity,
acceleration, and time. The kinematic equation to use is:
vf=vi+a·t
Step 3: Substitute the known values into the equation. Substitute the values
of vi,a,vf, and tinto the kinematic equation:
20 = 0 + 3.0·t
Step 4: Solve for the time taken. Solving for t:
20 = 3.0t
t=20
3.0
t= 6.7 seconds
Therefore, it will take the car approximately 6.7 seconds to reach a speed of
20 m/s.
Question 33
Question
A car accelerates from rest along a straight road with a constant acceleration
for 10 seconds until it reaches a velocity of 30 m/s. After maintaining this
velocity for another 5 seconds, the car decelerates uniformly until it comes to a
stop. If the total distance travelled by the car is 600 meters, determine: (a) the
acceleration of the car while accelerating, (b) the deceleration of the car while
coming to a stop, and (c) the total time taken for the car to come to a stop.
28
Solution
(a) First, we need to find the acceleration of the car while accelerating. Let
aacceleration be the acceleration of the car while accelerating. The car starts
from rest, so its initial velocity, u, is 0 m/s. The final velocity, v, is 30 m/s.
The time taken, tacceleration, is 10 seconds.
We can use the equation of motion that relates initial velocity, final velocity,
acceleration, and time:
v=u+at
Substitute u= 0, v= 30 m/s, and t= 10 s into the equation:
30 = 0 + a×10
a=30
10 = 3 m/s2
Therefore, the acceleration of the car while accelerating is 3 m/s2.
(b) Next, we need to find the deceleration of the car while coming to a stop.
Let adeceleration be the deceleration of the car. The car is initially moving with
a velocity of 30 m/s and comes to a stop. The final velocity, v, is 0 m/s. The
constant deceleration, adeceleration, is what we need to find. Let the time taken
for deceleration be tdeceleration.
We can use the kinematic equation:
v=u+at
Substitute u= 30 m/s, v= 0 m/s, and t=tdeceleration into the equation:
0 = 30 + adeceleration ×tdeceleration
adeceleration =−30
tdeceleration
(c) To find the time taken for the car to come to a stop, we can use the
equation for distance:
s=ut +1
2at2
For the deceleration phase, the distance covered is 300 meters. So we have:
300 = 30tdeceleration −1
230
tdeceleration tdeceleration2
300 = 30tdeceleration −15
15 = 30tdeceleration
tdeceleration =15
30 =1
2s
Therefore, the time taken for the car to come to a stop is 0.5 seconds.
29
Question 34
Question
A sprinter completes a 200 m race in 20.0 seconds. Starting from rest, what is
the sprinter’s average velocity during the race?
Solution
Step 1: Calculate the sprinter’s average velocity. The average velocity is given
by the displacement divided by the time taken:
Average Velocity = Displacement
Time Taken
Step 2: Find the displacement of the sprinter during the race. Since the
sprinter starts from rest and completes the race, the displacement is equal to
the distance of the race:
Displacement = 200 m
Step 3: Calculate the average velocity. Plugging in the values for displace-
ment and time into the formula for average velocity:
Average Velocity = 200 m
20.0 s
Average Velocity = 10.0 m/s
Therefore, the sprinter’s average velocity during the race is 10.0 m/s .
Question 35
Question
An object moves along the x-axis such that its position at time tis given by
x(t)=4t3−12t2+ 8t. Determine the velocity and acceleration of the object as
functions of time.
Solution
Step 1: To find the velocity, we need to differentiate the position function x(t)
with respect to time t.
Velocity v(t) = dx
dt =d
dt(4t3−12t2+ 8t)
Step 2: Differentiating each term separately:
= 12t2−24t+ 8
30
Solution
Step 1: Identify the knowns and unknowns.
The initial velocity (v0) is 0 m/s, the final velocity (v) is 25 m/s, and the time
(t) is 10 seconds. We need to find the acceleration (a).
Step 2: Choose a kinematic equation to use.
We can use the kinematic equation:
v=v0+at
Step 3: Plug in the known values.
Substitute v= 25 m/s, v0= 0 m/s, and t= 10 s into the kinematic equation:
25 = 0 + a(10)
Step 4: Solve for acceleration.
Solving for a, we get:
a=25
10
Step 5: Calculate the acceleration.
a= 2.5 m/s2
Step 6: Check the units.
The units for acceleration are in meters per second squared, which is the correct
unit for acceleration.
Question 3
Question
A car accelerates uniformly from rest to a speed of 25 m/s over a distance of
200 m. What is the acceleration of the car?
Solution
Step 1: Identify the given values: The initial velocity (vi) of the car is 0 m/s.
The final velocity (vf) of the car is 25 m/s. The displacement (d) of the car is
200 m.
Step 2: Use the kinematic equation relating initial velocity, final velocity,
acceleration, and displacement:
v2
f=v2
i+ 2ad
Step 3: Substitute the known values into the equation:
(25 m/s)2= (0 m/s)2+ 2a×200 m
2
Step 4: Simplify the equation:
625 = 400a
Step 5: Solve for the acceleration a:
a=625
400 = 1.5625 m/s2
Therefore, the acceleration of the car is 1.5625 m/s2.
Question 4
Question
A car accelerates from rest at a constant rate of 3 m/s2for 10 seconds. After this
time, the car maintains a constant speed for 20 seconds, and then decelerates
at a rate of 2 m/s2until it comes to a stop. What is the total distance traveled
by the car during this entire motion?
Solution
Step 1: Find the distance traveled during acceleration phase.
Time taken during acceleration phase, t1= 10 s
Acceleration, a= 3 m/s2
Initial velocity, u= 0 m/s
Using the kinematic equation s=ut +1
2at2, we find the distance traveled
during acceleration as:
s1= 0 + 1
2·3·(10)2= 150 m
Step 2: Find the distance traveled during constant speed phase.
Time taken during constant speed phase, t2= 20 s
Speed, v= 3 m/s
Distance traveled during constant speed phase is given by:
s2=v·t2= 3 ·20 = 60 m
Step 3: Find the distance traveled during deceleration phase.
Time taken during deceleration phase, t3= 10 s
Deceleration, a= 2 m/s2
3
Speed before deceleration, v= 3 m/s
Using the kinematic equation v2=u2+ 2as, where v= 0 as the car stops
at the end:
0=32+ 2 ·(−2) ·s
Solving for s, we get:
s= 2.25 m
Step 4: Calculate the total distance traveled by the car.
Total distance = s1+s2+s3= 150 + 60 + 2.25 = 212.25 m
Therefore, the total distance traveled by the car during this entire motion is
212.25 meters.
Question 5
Question
A particle moves along a straight line such that its position is given by x(t) =
5t3−2t2+ 3, where xis in meters and tis in seconds. Determine the particle’s
velocity and acceleration as functions of time.
Solution
Step 1: To find the particle’s velocity, we need to take the derivative of the
position function with respect to time (t).
Velocity v(t) = dx
dt
Step 2: Calculate the derivative of x(t):
v(t) = d(5t3−2t2+ 3)
dt
Step 3: Differentiate each term separately:
v(t) = d(5t3)
dt −d(2t2)
dt +d(3)
dt
Step 4: Find the derivatives of each term:
v(t) = 15t2−4t
Step 5: Therefore, the particle’s velocity as a function of time is v(t) =
15t2−4tm/s.
4
Step 6: To find the particle’s acceleration, we need to take the derivative of
the velocity function with respect to time.
Acceleration a(t) = dv
dt
Step 7: Calculate the derivative of v(t):
a(t) = d(15t2−4t)
dt
Step 8: Differentiate each term separately:
a(t) = d(15t2)
dt −d(4t)
dt
Step 9: Find the derivatives of each term:
a(t) = 30t−4
Step 10: Therefore, the particle’s acceleration as a function of time is a(t) =
30t−4 m/s2.
Question 6
Question
A car accelerates from rest at a constant rate of 2 m/s2. How fast is the car
going after 5 seconds?
Solution
Step 1: First, we need to determine the final velocity of the car using the
kinematic equation:
v=u+at
where v= final velocity, u= initial velocity, a= acceleration, and t= time.
Given that u= 0 (the car starts from rest), a= 2 m/s2, and t= 5 s, we can
plug these values into the equation:
v= 0 + 2 ×5
v= 10 m/s
Therefore, the car is going 10 m/s after 5 seconds.
5
Question 7
Question
A car accelerates uniformly from rest at 2.5 m/s2for 5 seconds, then continues
at a constant velocity for 10 seconds, and finally decelerates uniformly to come
to a stop in 4 seconds. Determine the total distance traveled by the car during
this time.
Solution
Step 1: Calculate the distance traveled during acceleration phase. Given that
initial velocity (vi) is 0 m/s, acceleration (a) is 2.5 m/s2, and time (t) is 5
seconds. We can use the kinematic equation: d=vit+1
2at2. Substitute the
values and calculate:
d= (0 ×5) + 1
2×2.5×52
d= 0 + 1
2×2.5×25
d= 0 + 31.25
d= 31.25 m
Step 2: Calculate the distance traveled during constant velocity phase. Since
the car is traveling at a constant velocity, the distance traveled is simply the
product of velocity and time. Given that the velocity (v) during this phase is
2.5 m/s and time (t) is 10 seconds:
d=v×t
d= 2.5×10
d= 25 m
Step 3: Calculate the distance traveled during deceleration phase. The car
comes to a stop during this phase, so the distance traveled can be calculated
using the same formula we used for acceleration. The final velocity (vf) is 0
m/s, acceleration (a) is -2.5 m/s2, and time (t) is 4 seconds:
d=vit+1
2at2
d= 0 ×4 + 1
2×(−2.5) ×42
d= 0 + 1
2×(−2.5) ×16
d= 0 −20
d=−20 m
6
Step 4: Calculate the total distance traveled by the car. The total distance
traveled is the sum of the distances calculated in each phase. Total distance =
31.25 m + 25 m - 20 m Total distance = 36.25 m
Therefore, the total distance traveled by the car during this time is 36.25
meters.
Question 8
Question
A car is traveling along a straight road. The car accelerates from rest at a
constant rate of 2 m/s2for 10 seconds, then maintains a constant speed for 20
seconds, and finally decelerates at a rate of −3 m/s2until it comes to a stop.
Determine the total distance the car travels during this time interval.
Solution
Step 1: Calculate the distance traveled during the acceleration phase. The
distance traveled during constant acceleration can be calculated using the equa-
tion:
d=1
2at2
where ais the acceleration and tis the time interval.
Given a= 2 m/s2and t= 10 s:
d=1
2×2 m/s2×(10 s)2
d= 100 m
Step 2: Calculate the distance traveled during the constant speed phase.
Since the speed is constant, the distance traveled is given by:
distance = speed ×time
Given speed = 20 m/s and time = 20 s:
distance = 20 m/s ×20 s
distance = 400 m
Step 3: Calculate the distance traveled during the deceleration phase. The
distance traveled during deceleration can be calculated using the equation:
d=vit+1
2at2
where viis the initial velocity, tis the time interval, and ais the acceleration.
7
Given vi= 20 m/s, a=−3 m/s2, and t= 10 s:
d= 20 m/s ×10 s + 1
2× −3 m/s2×(10 s)2
d= 200 m −150 m
d= 50 m
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of the distances during each phase:
Total distance = 100 m + 400 m + 50 m
Total distance = 550 m
Therefore, the total distance the car travels during this time interval is 550
meters.
Question 9
Question
A car is traveling along a straight road. Initially, the car is moving with a
velocity of 15 m/s, and then it accelerates at a constant rate of 2 m/s2for 8
seconds. After this time, the car brakes and decelerates at a constant rate of
3 m/s2until it comes to a stop. Determine the total distance traveled by the
car during this entire motion.
Solution
Step 1: Find the distance covered during acceleration phase. Using the equation
of motion: v=u+at, where vis the final velocity, uis the initial velocity, a
is the acceleration, and tis the time, the final velocity after 8 seconds can be
found as: v= 15 + 2(8) = 31 m/s.
The distance covered during acceleration phase can be calculated using the
formula: s=ut +1
2at2, where sis the distance, uis the initial velocity, ais the
acceleration, and tis the time, s= 15(8) + 1
2·2·(8)2= 120 + 64 = 184 m.
Step 2: Find the distance covered during deceleration phase. The final
velocity after decelerating to stop can be found using the equation of motion:
v2=u2+ 2as, where vis the final velocity (0 in this case), uis the initial
velocity (31 m/s), ais the acceleration (3 m/s2), and sis the distance, 0 =
(31)2+ 2 ·(−3) ·s. Solving for sgives: s=(31)2
2·3=961
6= 160.17 m.
Step 3: Find the total distance traveled. The total distance traveled by
the car is the sum of the distances covered during acceleration and deceleration
phases: Total distance = 184 m + 160.17 m = 344.17 m.
Therefore, the total distance traveled by the car during its entire motion is
344.17 meters.
8
Question 10
Question
A ball is thrown vertically upward with an initial velocity of 30 m/s. How long
does it take for the ball to reach its maximum height? What is the maximum
height the ball reaches? (Assume the acceleration due to gravity is 9.81 m/s2)
Solution
Step 1: Identify the known variables and the equation to be used.
Initial velocity (vi) = 30 m/s
Acceleration due to gravity (g) = 9.81 m/s2
Final velocity at the maximum height (vf) = 0 m/s (ball momentarily
stops at the maximum height)
We will use the kinematic equation for motion with constant acceleration:
vf=vi+at where vf= 0.
Step 2: Find the time it takes for the ball to reach the maximum height.
Using the kinematic equation vf=vi+at with vf= 0:
0 = 30 + (−9.81)t
9.81t= 30
t=30
9.81 ≈3.06 s
Step 3: Find the maximum height the ball reaches. We can use the kinematic
equation for displacement: s=vit+1
2at2. Substitute vi= 30 m/s, t= 3.06 s,
and a=−9.81 m/s2into the equation:
s= 30(3.06) + 1
2(−9.81)(3.06)2
s= 91.8−45.2
s≈46.6 m
Therefore, it takes approximately 3.06 seconds for the ball to reach its max-
imum height, and the maximum height the ball reaches is approximately 46.6
meters.
Question 11
Question
A car initially at rest starts moving in a straight line with a constant acceleration
of 3 m/s2. How long will it take for the car to reach a speed of 30 m/s?
9
Solution
Step 1: Let’s denote the initial velocity of the car as v0, the final velocity as vf,
the acceleration as a, and the time taken to reach the final velocity as t. Given:
v0= 0, a= 3 m/s2, and vf= 30 m/s.
Step 2: We can use the kinematic equation:
vf=v0+a·t
Substitute the known values:
30 = 0 + 3t
Step 3: Solve for t:
t=30
3= 10 seconds
Step 4: Therefore, it will take the car 10 seconds to reach a speed of 30 m/s.
Question 12
Question
A car is traveling along a straight road. The car starts from rest and accelerates
uniformly at 2.5 m/s2for 10 seconds. After this time, the car maintains a
constant velocity for another 20 seconds. Finally, the car decelerates uniformly
to a stop in 5 seconds. Calculate the total distance the car travels during this
entire period.
Solution
Step 1: Calculate the distance traveled during acceleration phase. The distance
traveled during acceleration can be calculated using the kinematic equation:
d=1
2·a·t2
Substitute a= 2.5 m/s2and t= 10 s into the equation:
d=1
2·2.5 m/s2·(10 s)2
d=1
2·2.5 m/s2·100 s2
d= 125 m
So, the car travels 125 meters during the acceleration phase.
Step 2: Calculate the distance traveled during constant velocity phase. Dur-
ing the constant velocity phase, the car moves at a constant speed, which can
be calculated by multiplying the constant velocity by the time:
d=v·t
10
Substitute the constant velocity and the time:
d= 25 m/s ·20 s
d= 500 m
So, the car travels 500 meters during the constant speed phase.
Step 3: Calculate the distance traveled during deceleration phase. The dis-
tance traveled during the deceleration phase can be calculated using the kine-
matic equation:
d=1
2·a·t2
Substitute a=−2.5 m/s2(as it’s deceleration) and t= 5 s into the equation:
d=1
2·(−2.5) m/s2·(5 s)2
d=1
2·(−2.5) m/s2·25 s2
d=−31.25 m
So, the car travels 31.25 meters during the deceleration phase.
Step 4: Calculate the total distance traveled by the car. To calculate the
total distance, we sum up the distances traveled during each phase:
T otal distance = 125 m + 500 m + 31.25 m
T otal distance = 656.25 m
Therefore, the car travels a total distance of 656.25 meters.
Question 13
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2for 10
seconds. After this time, the car maintains a constant speed for the next 20
seconds. Finally, the car decelerates at a rate of 1.5 m/s2until it comes to a
stop. Determine the total distance traveled by the car during this entire trip.
Solution
Step 1: Find the distance traveled during acceleration phase.
The distance traveled during acceleration is given by the formula:
d=1
2×acceleration ×time2
11
Substitute the values: acceleration = 2.5 m/s2, time = 10 seconds.
d=1
2×2.5×102
d=1
2×2.5×100
d= 125 m
Therefore, the distance traveled during the acceleration phase is 125 m.
Step 2: Find the distance traveled during constant speed phase.
During constant speed, the distance traveled is given by the formula:
d= speed ×time
We know the car maintained a speed, so the distance traveled during this
phase is:
d= speed ×time
d= speed ×20
Step 3: Find speed during constant speed phase.
Since the car maintained a constant speed during the 20 seconds, the speed
is the final speed at the end of the acceleration phase. The final speed is given
by:
Final speed = initial speed + acceleration ×time
Substitute the values: initial speed = 0 m/s, acceleration = 2.5 m/s2, time
= 10 seconds.
Final speed = 0 + 2.5×10
Final speed = 0 + 25
Final speed = 25 m/s
Therefore, the speed during the constant speed phase is 25 m/s.
Step 4: Calculate the distance traveled during constant speed phase.
d= speed ×20
d= 25 ×20
d= 500 m
Therefore, the distance traveled during the constant speed phase is 500 m.
Step 5: Find the distance traveled during the deceleration phase.
The distance traveled during deceleration is given by the formula:
d=1
2×deceleration ×time2
Substitute the values: deceleration = 1.5 m/s2, time = unknown, final speed
= 0 m/s.
0 = 25 + (−1.5) ×time
12
25 = 1.5×time
time = 25
1.5
time = 16.67 seconds
Therefore, the total time for deceleration is approximately 16.67 seconds.
Now, substitute the time into the distance formula:
d=1
2×1.5×(16.67)2
d=1
2×1.5×278.89
d= 208.67 m
Thus, the distance traveled during the deceleration phase is 208.67 m.
Step 6: Calculate the total distance traveled by the car.
The total distance traveled is the sum of the distances traveled during each
phase:
Total distance = 125 + 500 + 208.67
Total distance = 833.67 m
Therefore, the total distance traveled by the car during this entire trip is
approximately 833.67 meters.
Question 14
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the car continues at a constant velocity for another 20 seconds.
If the total distance traveled by the car during this time is 500 meters, find the
distance traveled during the constant velocity phase.
Solution
Step 1: Find the distance traveled during the acceleration phase. To find the
distance traveled during the acceleration phase, we first need to find the final
velocity of the car after accelerating for 10 seconds. We can use the equation
of motion: v=u+at, where: - vis the final velocity, - uis the initial velocity
(which is 0 m/s since the car starts from rest), - ais the acceleration of the
car (which is 2 m/s2), and - tis the time the car accelerates for (which is 10
seconds).
Substitute the values into the equation to find v:
v= 0 + (2 m/s2)(10 s) = 20 m/s
13
Next, use the formula for distance traveled during constant acceleration mo-
tion: s=ut +1
2at2, where: - sis the distance traveled, - uis the initial velocity,
-tis the time, and - ais the acceleration.
Substitute the values into the formula to find the distance traveled during
the acceleration phase:
sacc = (0 m/s)(10 s) + 1
2(2 m/s2)(10 s)2= 100 m
So, the distance traveled during the acceleration phase is 100 meters.
Step 2: Find the distance traveled during the constant velocity phase. Since
the total distance traveled by the car is 500 meters and the distance traveled
during the acceleration phase is 100 meters, the distance traveled during the
constant velocity phase can be found by subtraction:
sconst = 500 m −100 m = 400 m
Therefore, the distance traveled during the constant velocity phase is 400
meters.
Question 15
Question
A car accelerates from rest at a constant rate of 2.5 m/s2for 10 seconds. After
this time, what is its displacement from the starting point?
Solution
Step 1: Identify the given variables. Given: Initial velocity, u= 0 m/s (as the
car starts from rest); Acceleration, a= 2.5 m/s2; Time, t= 10 s.
Step 2: Use the kinematic equation for displacement. The equation for
displacement in terms of initial velocity, acceleration, and time is:
s=ut +1
2at2
Step 3: Substitute the given values into the equation.
s= (0)(10) + 1
2(2.5)(10)2
Step 4: Calculate the displacement.
s= 0 + 1
2(2.5)(100)
s=1
2×250
s= 125 meters
Step 5: Answer After accelerating at 2.5 m/s2for 10 seconds, the car’s
displacement from the starting point is 125 meters.
14
Question 16
Question
A car accelerates uniformly from rest at 2.0 m/s2. How far will the car have
traveled when it reaches a velocity of 25 m/s?
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, u= 0 m/s (rest) Acceleration, a= 2.0 m/s2Final
velocity, v= 25 m/s Distance traveled, x= ?
Step 2: Determine the time taken to reach the final velocity.
The final velocity can be related to the initial velocity, acceleration, and time
using the kinematic equation v=u+at, where:
v=u+at
25 = 0 + 2t
t=25
2
t= 12.5 seconds
Step 3: Calculate the distance traveled using the equation x=ut +1
2at2.
Substitute the known values into the equation:
x= 0(12.5) + 1
2(2)(12.5)2
x= 0 + 0.5(2)(156.25)
x= 0 + 156.25
x= 156.25 meters
Therefore, the car will have traveled 156.25 meters when it reaches a velocity
of 25 m/s.
Question 17
Question
A car starts from rest at a traffic light and accelerates uniformly at 3.0 m/s2for
8.0 seconds. After this, the car continues at a constant velocity for an additional
10 seconds. Determine the total distance the car travels during this time period.
15
Solution
Step 1: Find the distance traveled during the acceleration phase. The distance
traveled during the acceleration phase can be found using the equation:
d1=1
2at2
1
where: a= 3.0 m/s2(acceleration), t1= 8.0 s (time during acceleration).
Substitute the values and solve for d1:
d1=1
2×3.0 m/s2×(8.0 s)2
d1= 0.5×3.0 m/s2×64.0 s2
d1= 96.0 m
Therefore, the distance traveled during the acceleration phase is 96.0 meters.
Step 2: Find the distance traveled during the constant velocity phase. Dur-
ing the constant velocity phase, the distance traveled is given by:
d2=v×t2
where: vis the constant velocity, t2= 10 s (time during constant velocity).
Since the car continued at a constant velocity after the acceleration phase,
the final velocity at the end of the acceleration phase is the velocity during the
constant velocity phase. We can find the final velocity using:
vf=a×t1
vf= 3.0 m/s2×8.0 s
vf= 24.0 m/s
Substitute the values to find d2:
d2= 24.0 m/s ×10 s
d2= 240.0 m
Therefore, the distance traveled during the constant velocity phase is 240.0
meters.
Step 3: Find the total distance traveled. The total distance traveled is
the sum of the distances traveled during the acceleration and constant velocity
phases:
Total distance = d1+d2
Total distance = 96.0 m + 240.0 m
Total distance = 336.0 m
Hence, the total distance the car travels during this time period is 336.0
meters.
16
Question 18
Question
A car accelerates uniformly from rest to a speed of 30 m/s over a distance of
150 m. What is the magnitude of the car’s acceleration?
Solution
Step 1: Identify the given information.
The final velocity of the car, vf, is 30 m/s. The initial velocity, vi, is 0 m/s
(since the car starts from rest). The displacement of the car, ∆x, is 150 m.
Step 2: Determine the acceleration using the kinematic equation:
v2
f=v2
i+ 2a∆x
Step 3: Plug in the known values and solve for acceleration:
a=v2
f−v2
i
2∆x
a=(30 m/s)2−(0 m/s)2
2(150 m)
a=900 m2/s2
300 m
a= 3 m/s2
Therefore, the magnitude of the car’s acceleration is 3 m/s2.
Question 19
Question
A car accelerates from rest at a constant rate of 2.0 m/s2along a straight road.
After 5.0 seconds, what is the displacement of the car?
Solution
Step 1: First, we need to determine the final velocity of the car after 5.0 seconds
using the kinematic equation:
v=v0+at
where vis the final velocity, v0is the initial velocity, ais the acceleration, and t
is the time. Given that the initial velocity v0= 0 m/s, the acceleration a= 2.0
m/s2, and the time t= 5.0 s, we have:
v= 0 + (2.0 m/s2)(5.0 s) = 10 m/s
17
Step 2: Next, we can find the displacement of the car using the kinematic
equation:
s=v0t+1
2at2
where sis the displacement, v0is the initial velocity, ais the acceleration, and t
is the time. Given that the initial velocity v0= 0 m/s, the acceleration a= 2.0
m/s2, and the time t= 5.0 s, we have:
s= 0(5.0 s) + 1
2(2.0 m/s2)(5.0 s)2= 25 m
Therefore, the displacement of the car after 5.0 seconds is 25 meters.
Question 20
Question
A car is traveling along a straight road at a constant velocity of 25 m/s for 20
seconds. It then accelerates uniformly to a velocity of 40 m/s over the next 10
seconds. Finally, it maintains this velocity for an additional 30 seconds before
coming to a stop with a constant deceleration. Find the total distance the car
traveled during this entire time.
Solution
Step 1: Find the distance traveled during the first phase of motion (constant
velocity). The distance traveled during constant velocity can be found using
the formula:
d=v×t
where dis the distance, vis the velocity, and tis the time. Given v= 25 m/s
and t= 20 seconds:
d= 25 ×20 = 500 meters
Step 2: Find the distance traveled during the second phase of motion (uni-
form acceleration). The distance traveled during uniform acceleration can be
found using the formula:
d=vi×t+1
2×a×t2
where dis the distance, viis the initial velocity, ais the acceleration, and tis
the time. Given vi= 25 m/s, a=40−25
10 = 1.5 m/s2, and t= 10 seconds:
d= 25 ×10 + 1
2×1.5×(10)2= 250 + 75 = 325 meters
Step 3: Find the distance traveled during the third phase of motion (constant
velocity). Similar to Step 1: Given v= 40 m/s and t= 30 seconds:
d= 40 ×30 = 1200 meters
18
Step 4: Find the distance traveled during the fourth phase of motion (de-
celeration). Since the car comes to a stop with constant deceleration, the total
distance traveled during deceleration is equal to the distance traveled in the
third phase (since the car comes to a stop). Therefore, the distance traveled
during deceleration is 1200 meters.
Step 5: Add up the distances from each phase to find the total distance
traveled. The total distance traveled is the sum of the distances from each
phase:
500 + 325 + 1200 + 1200 = 3225 meters
Question 21
Question
A car accelerates from rest at a constant rate of 3 m/s2for 10 seconds. After
this time, the car slows down with a constant acceleration of −2 m/s2until it
comes to a stop. Determine the total distance travelled by the car during this
motion.
Solution
Step 1: Find the distance travelled during the acceleration phase. The distance
travelled during acceleration can be found using the kinematic equation:
d1=vit+1
2a1t2
where - d1is the distance travelled during acceleration, - viis the initial velocity
during acceleration (which is 0 since the car starts from rest), - a1is the accel-
eration during the acceleration phase (which is 3 m/s2), - tis the time during
which the car accelerates (which is 10 seconds).
Plugging in the values, we get:
d1= 0 ×10 + 1
2×3×(10)2
d1= 0 + 1
2×3×100
d1= 0 + 1.5×100
d1= 150 m
Step 2: Find the distance travelled during the deceleration phase. The
distance travelled during deceleration can also be found using the kinematic
equation, however, since the car comes to a stop, the final velocity during de-
celeration is 0. Therefore, the distance travelled during deceleration can be
expressed as:
d2=v2
f−v2
i
2a2
19
where - d2is the distance travelled during deceleration, - vfis the final velocity
during deceleration (which is 0), - viis the initial velocity during deceleration,
-a2is the acceleration during the deceleration phase (which is −2 m/s2).
During the acceleration phase, the final velocity was:
vi=vi+a1t= 0 + 3 ×10 = 30 m/s
Plugging in the values, we get:
d2=(0)2−(30)2
2× −2
d2=0−900
−4
d2=−900
−4
d2= 225 m
Step 3: Find the total distance travelled by the car. The total distance
travelled by the car is the sum of the distances travelled during acceleration
and deceleration:
Total distance = d1+d2= 150 m + 225 m = 375 m
Therefore, the total distance traveled by the car during this motion is 375 m .
Question 22
Question
A car is initially traveling at v0= 20 m/s and accelerates uniformly at a=
3 m/s2for a distance of d= 100 m. What is the final velocity of the car?
Solution
Step 1: Calculate the final velocity using the equation of motion for uniformly
accelerated motion:
Final velocity2= Initial velocity2+ 2 ×acceleration ×distance
Final velocity = qv2
0+ 2ad
Step 2: Substitute the given values into the equation:
Final velocity = q(20 m/s)2+ 2 ×(3 m/s2)×(100 m)
Step 3: Perform the calculation:
Final velocity = √400 + 600
Final velocity = √1000
Final velocity = 31.62 m/s
Therefore, the final velocity of the car is 31.62 m/s.
20
Question 23
Question
A car accelerates from rest at a rate of 3 m/s2for 10 seconds. After this time,
the car maintains a constant velocity for an additional 20 seconds. Finally, the
car decelerates at a rate of 2 m/s2until it comes to a stop. Calculate the total
distance traveled by the car during this entire process.
Solution
Step 1: Find the distance traveled during the acceleration phase. The distance
traveled during acceleration can be found using the equation:
d=1
2at2
where dis the distance, ais the acceleration, and tis the time. Plugging in the
values, we get:
d=1
2×3×(10)2
d=1
2×3×100
d= 150 m
Step 2: Find the distance traveled during the constant velocity phase. Since
the velocity is constant, the distance traveled can be calculated using the equa-
tion:
d=vt
where dis the distance, vis the velocity, and tis the time. Given that the car
maintains a constant velocity for 20 seconds, the distance traveled during this
phase is:
d=v×20
Since the initial velocity is 30 m/s (acquired during the acceleration phase), the
distance traveled during this phase is:
d= 30 ×20
d= 600 m
Step 3: Find the distance traveled during the deceleration phase. The dis-
tance traveled during deceleration can be found using the equation:
d=vit+1
2at2
where dis the distance, viis the initial velocity, ais the acceleration, and t
is the time. Given that the initial velocity is 30 m/s (as the car maintains a
21
constant velocity of 30 m/s for 20 seconds) and the deceleration rate is −2 m/s2,
the distance traveled during this phase is:
d= 30 ×20 + 1
2×(−2) ×(20)2
d= 600 −200
d= 400 m
Step 4: Calculate the total distance traveled. The total distance traveled by
the car is the sum of the distances traveled in each phase:
Total distance = 150 + 600 + 400
Total distance = 1150 m
Therefore, the total distance traveled by the car during this process is 1150
meters.
Question 24
Question
A car starts from rest and accelerates at a constant rate of 2.0 m/s2. How fast
is it going after it has traveled 100 meters? (The car starts at x= 0, with initial
velocity v0= 0, and acceleration a= 2.0 m/s2)
Solution
Step 1: Use the kinematic equation for velocity to find the final velocity of the
car.
v=v0+at
Since v0= 0 and a= 2.0 m/s2, we have
v= 0 + 2.0 m/s2·t
Step 2: Find the time ttaken for the car to travel 100 meters using the
kinematic equation for position.
x=x0+v0t+1
2at2
Since x= 100 m, x0= 0, v0= 0, and a= 2.0 m/s2, we have
100 = 0 + 0 + 1
2·2.0 m/s2·t2
Solving for t:
100 = t2
22
t= 10 s
Step 3: Substitute t= 10 s into the expression for velocity to find the final
velocity v.
v= 2.0 m/s2·10 s
v= 20 m/s
Answer: The car is going 20 m/s after traveling 100 meters.
Question 25
Question
A car accelerates from rest at a constant rate of 2 m/s2for a distance of 50
meters before reaching a constant velocity. How long does it take for the car to
reach this constant velocity?
Solution
Step 1: We can find the time it takes for the car to reach its constant velocity
using the kinematic equation for motion in one dimension:
v=u+at
where: - vis the final velocity, - uis the initial velocity, - ais the acceleration,
and - tis the time taken.
Step 2: Since the car starts from rest, the initial velocity u= 0 m/s.
Step 3: For the accelerating phase, we are given:
a= 2 m/s2
u= 0 m/s
v=?
Step 4: Using the equation v=u+at, we can solve for v:
v= 0 + (2)(t)
v= 2t
Step 5: During the acceleration phase, the car covers a distance of 50 meters.
We know that distance = average velocity ×time.
Step 6: The average velocity during acceleration is given by:
Average velocity = u+v
2
Average velocity = 0+2t
2=t
23
Step 7: Since the car travels 50 meters during the acceleration phase:
50 = t×t
50 = t2
Step 8: Solving for t:
t=√50
t≈7.07 s
Step 9: Therefore, it takes approximately 7.07 seconds for the car to reach
its constant velocity.
Question 26
Question
A car initially traveling at a speed of 20 m/s undergoes constant acceleration
and comes to a stop in 100 meters. What is the magnitude of the acceleration
of the car?
Solution
Step 1: Identify the known quantities. The initial velocity of the car, u, is 20
m/s, the final velocity of the car, v, is 0 m/s, the displacement of the car, s, is
100 meters.
Step 2: Choose the appropriate kinematic equation. We can use the equation
that relates initial velocity, final velocity, acceleration, and displacement without
time:
v2=u2+ 2as
Step 3: Plug in the known values and solve for the acceleration. Substitute
v,u, and sinto the equation:
0 = (20)2+ 2a(100)
0 = 400 + 200a
Solving for a:
a=−400
200 =−2 m/s2
Step 4: State the answer. The magnitude of the acceleration of the car is 2
m/s2.
24
Question 27
Question
A particle moves along a straight line according to the equation x= 4t2−3t+5,
where xis in meters and tis in seconds. Find the acceleration of the particle
when t= 3 s.
Solution
Step 1: Find the velocity of the particle using the derivative of the position
function.
v(t) = dx
dt =d
dt(4t2−3t+ 5)
v(t)=8t−3
Step 2: Find the acceleration of the particle using the derivative of the
velocity function.
a(t) = dv
dt =d
dt(8t−3)
a(t)=8
Step 3: Calculate the acceleration of the particle at t= 3 s.
a(3) = 8 m/s2
Therefore, the acceleration of the particle at t= 3 s is 8 m/s2.
Question 28
Question
A car is initially traveling at a speed of 20 m/s. It then accelerates at a constant
rate of 3 m/s2for 8 seconds. What is the final velocity of the car after this
acceleration period?
Solution
Step 1: Determine the acceleration of the car using the equation a=∆v
∆t, where
ais the acceleration, ∆vis the change in velocity, and ∆tis the time interval.
Given: Initial velocity vi= 20 m/s, acceleration a= 3 m/s2, time interval
∆t= 8 s. Plug in the values: a=vf−vi
∆t3 = vf−20
824 = vf−20 vf= 24 + 20
vf= 44 m/s
Therefore, the final velocity of the car after accelerating at a constant rate
of 3 m/s2for 8 seconds is 44 m/s.
25
Question 29
Question
A car starts from rest and accelerates along a straight road according to the
equation x(t)=2t2+ 3t+ 1, where xis in meters and tis in seconds. Calculate
the acceleration of the car at t= 2 seconds.
Solution
Step 1: To find the acceleration of the car, we need to differentiate the displace-
ment function x(t) with respect to time tto get the velocity function, and then
differentiate the velocity function to get the acceleration function.
Step 2: Let’s start by finding the velocity function v(t) by differentiating
x(t) with respect to t:
v(t) = dx
dt =d
dt(2t2+ 3t+ 1) = 4t+ 3
Step 3: Next, let’s find the acceleration function a(t) by differentiating v(t)
with respect to t:
a(t) = dv
dt =d
dt(4t+ 3) = 4
Step 4: Now, to find the acceleration of the car at t= 2 seconds, substitute
t= 2 into the acceleration function:
a(2) = 4 m/s2
Therefore, the acceleration of the car at t= 2 seconds is 4 m/s2.
Question 30
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 10
seconds. After this time, the car maintains a constant velocity for an additional
20 seconds. Finally, the car decelerates at a rate of 2.0 m/s2until it comes to a
stop. Determine the total distance the car travels during this entire process.
Solution
Step 1: Determine the distance traveled during acceleration phase. Given that
the car starts from rest and accelerates at 3.0 m/s2for 10 seconds, we can use
the equation x=vit+1
2at2, where xis the distance traveled, viis the initial
velocity (0 m/s), ais the acceleration, and tis the time.
x= 0 ·10 + 1
2·3.0·(10)2= 150 m
26
Step 2: Determine the distance traveled during constant velocity phase.
During this phase, the car travels at a constant velocity for 20 seconds. Since
velocity is constant, distance traveled can be calculated by multiplying velocity
by time.
x= 3.0·20 = 60 m
Step 3: Determine the distance traveled during deceleration phase. In this
phase, the car decelerates at 2.0 m/s2until it comes to a stop. We can again
use the equation x=vit+1
2at2, where viis the initial velocity (3.0 m/s), ais
the deceleration (-2.0 m/s2), and tis the time. The time taken to stop can be
calculated as t=vf−vi
a=0−3.0
−2.0= 1.5 seconds. The distance traveled during
deceleration phase is then
x= 3.0·1.5 + 1
2·(−2.0) ·(1.5)2= 4.5−1.125 = 3.375 m
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of the distances traveled during each phase.
Total Distance = 150 + 60 + 3.375 = 213.375 m
Therefore, the total distance the car travels during this entire process is
213.375 meters.
Question 31
Question
A car starts from rest and accelerates at a constant rate of 3.50 m/s2. How long
does it take for the car to reach a speed of 28.0 m/s?
Solution
Step 1: We can use the kinematic equation to solve for the time it takes for the
car to reach a speed of 28.0 m/s. The kinematic equation relating final velocity,
initial velocity, acceleration, and time is given by:
v=u+at
where: v= final velocity (28.0 m/s), u= initial velocity (0 m/s, as the car
starts from rest), a= acceleration (3.50 m/s2), t= time.
Step 2: Substituting the given values into the kinematic equation, we have:
28.0 m/s = 0 m/s + 3.50 m/s2·t
Step 3: Solve for tby isolating it on one side of the equation:
t=28.0 m/s
3.50 m/s2
27
Step 4: Calculate the time it takes for the car to reach a speed of 28.0 m/s:
t=28.0
3.50 = 8.00 s
Answer: The car will take 8.00 seconds to reach a speed of 28.0 m/s.
Question 32
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2along a straight
road. How long does it take for the car to reach a speed of 20 m/s?
Solution
Step 1: Identify the given quantities and the unknown. Let: - The initial velocity
of the car be vi= 0 m/s (starting from rest). - The acceleration of the car be
a= 3.0 m/s2. - The final velocity of the car be vf= 20 m/s. - The time taken
for the car to reach the final velocity be t(unknown).
Step 2: Use the kinematic equation relating initial velocity, final velocity,
acceleration, and time. The kinematic equation to use is:
vf=vi+a·t
Step 3: Substitute the known values into the equation. Substitute the values
of vi,a,vf, and tinto the kinematic equation:
20 = 0 + 3.0·t
Step 4: Solve for the time taken. Solving for t:
20 = 3.0t
t=20
3.0
t= 6.7 seconds
Therefore, it will take the car approximately 6.7 seconds to reach a speed of
20 m/s.
Question 33
Question
A car accelerates from rest along a straight road with a constant acceleration
for 10 seconds until it reaches a velocity of 30 m/s. After maintaining this
velocity for another 5 seconds, the car decelerates uniformly until it comes to a
stop. If the total distance travelled by the car is 600 meters, determine: (a) the
acceleration of the car while accelerating, (b) the deceleration of the car while
coming to a stop, and (c) the total time taken for the car to come to a stop.
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Solution
(a) First, we need to find the acceleration of the car while accelerating. Let
aacceleration be the acceleration of the car while accelerating. The car starts
from rest, so its initial velocity, u, is 0 m/s. The final velocity, v, is 30 m/s.
The time taken, tacceleration, is 10 seconds.
We can use the equation of motion that relates initial velocity, final velocity,
acceleration, and time:
v=u+at
Substitute u= 0, v= 30 m/s, and t= 10 s into the equation:
30 = 0 + a×10
a=30
10 = 3 m/s2
Therefore, the acceleration of the car while accelerating is 3 m/s2.
(b) Next, we need to find the deceleration of the car while coming to a stop.
Let adeceleration be the deceleration of the car. The car is initially moving with
a velocity of 30 m/s and comes to a stop. The final velocity, v, is 0 m/s. The
constant deceleration, adeceleration, is what we need to find. Let the time taken
for deceleration be tdeceleration.
We can use the kinematic equation:
v=u+at
Substitute u= 30 m/s, v= 0 m/s, and t=tdeceleration into the equation:
0 = 30 + adeceleration ×tdeceleration
adeceleration =−30
tdeceleration
(c) To find the time taken for the car to come to a stop, we can use the
equation for distance:
s=ut +1
2at2
For the deceleration phase, the distance covered is 300 meters. So we have:
300 = 30tdeceleration −1
230
tdeceleration tdeceleration2
300 = 30tdeceleration −15
15 = 30tdeceleration
tdeceleration =15
30 =1
2s
Therefore, the time taken for the car to come to a stop is 0.5 seconds.
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Question 34
Question
A sprinter completes a 200 m race in 20.0 seconds. Starting from rest, what is
the sprinter’s average velocity during the race?
Solution
Step 1: Calculate the sprinter’s average velocity. The average velocity is given
by the displacement divided by the time taken:
Average Velocity = Displacement
Time Taken
Step 2: Find the displacement of the sprinter during the race. Since the
sprinter starts from rest and completes the race, the displacement is equal to
the distance of the race:
Displacement = 200 m
Step 3: Calculate the average velocity. Plugging in the values for displace-
ment and time into the formula for average velocity:
Average Velocity = 200 m
20.0 s
Average Velocity = 10.0 m/s
Therefore, the sprinter’s average velocity during the race is 10.0 m/s .
Question 35
Question
An object moves along the x-axis such that its position at time tis given by
x(t)=4t3−12t2+ 8t. Determine the velocity and acceleration of the object as
functions of time.
Solution
Step 1: To find the velocity, we need to differentiate the position function x(t)
with respect to time t.
Velocity v(t) = dx
dt =d
dt(4t3−12t2+ 8t)
Step 2: Differentiating each term separately:
= 12t2−24t+ 8
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Therefore, the velocity of the object as a function of time is v(t) = 12t2−
24t+ 8.
Step 3: To find the acceleration, we need to differentiate the velocity function
v(t) with respect to time t.
Acceleration a(t) = dv
dt =d
dt(12t2−24t+ 8)
Step 4: Differentiating each term separately:
= 24t−24
Therefore, the acceleration of the object as a function of time is a(t) =
24t−24.
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