PHYS 202 - GENERAL PHYSICS II -
Kinematics in One Dimension
Question Bank - Set 4
Liberty University
Question 1
Question
A car initially at rest accelerates uniformly along a straight road, reaching a
speed of 25 m/s in 10 seconds. What is the acceleration of the car?
Solution
Step 1: Identify the known values and the unknown. Let: - Initial velocity,
u= 0 m/s - Final velocity, v= 25 m/s - Time taken, t= 10 s - Acceleration,
a=?
Step 2: Use the kinematic equation relating final velocity, initial velocity,
acceleration, and time:
v=u+at
Substitute the known values into the equation:
25 = 0 + a(10)
Step 3: Solve for the acceleration.
a=25
10
a= 2.5 m/s2
Therefore, the acceleration of the car is 2.5 m/s2.
Question 2
Question
A car accelerates uniformly from rest at 2.0 m/s2along a straight road. How
long does it take for the car to reach a speed of 25 m/s?
Solution
Step 1: Identify the given variables and the unknown variable. Let a= 2.0 m/s2
be the acceleration of the car, vf= 25 m/s be the final velocity, and tbe the
time taken for the car to reach a speed of 25 m/s.
Step 2: Find the equation connecting the variables. We know that the final
velocity of an object accelerated from rest is given by the equation:
vf=at
Step 3: Substitute the known values into the equation. Substitute vf= 25
m/s and a= 2.0 m/s2into the equation:
25 = 2.0t
Step 4: Solve for the unknown variable. Solving for t, we get:
t=25
2.0= 12.5 s
Therefore, it takes 12.5 seconds for the car to reach a speed of 25 m/s.
Question 3
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the car maintains a constant velocity for 20 seconds before
coming to a stop with a constant deceleration of 3 m/s2. Calculate the total
distance traveled by the car during this entire motion.
Solution
Step 1: Find the distance traveled during acceleration phase. The distance
covered during acceleration can be calculated using the equation:
s=ut +1
2at2
where uis the initial velocity, ais the acceleration, tis the time.
Given that the initial velocity, u= 0 m/s, acceleration, a= 2 m/s2, and
time, t= 10 s, we can plug these values into the equation to find the distance
traveled during acceleration:
s= 0 ×10 + 1
2×2×(10)2
s= 0 + 1
2×2×100
2
s= 0 + 1 ×100
s= 100 m
Therefore, the distance covered during acceleration phase is 100 m.
Step 2: Find the distance traveled during constant velocity phase. Since the
car maintains a constant velocity during this phase, the distance traveled can
be calculated using the equation:
s=vt
where vis the constant velocity and tis the time.
Given that the constant velocity, v= 2 ×10 = 20 m/s, and time, t= 20 s,
we can find the distance traveled during constant velocity motion:
s= 20 ×20
s= 400 m
Therefore, the distance covered during the constant velocity phase is 400 m.
Step 3: Find the distance traveled during deceleration phase. The distance
covered during deceleration can be calculated using the same equation as for
acceleration:
s=vt +1
2at2
where vis the final velocity, ais the deceleration, tis the time.
Since the car comes to a stop at the end of the deceleration phase, the final
velocity, v= 0 m/s, deceleration, a= 3 m/s2, and time, t= 20 s, we can plug
these values into the equation to find the distance traveled during deceleration:
s= 0 ×20 + 1
2×3×(20)2
s= 0 + 1
2×3×400
s= 0 + 1.5×400
s= 600 m
Therefore, the distance covered during the deceleration phase is 600 m.
Step 4: Calculate the total distance traveled by the car. The total distance
traveled by the car is the sum of distances traveled during acceleration, constant
velocity, and deceleration phases: Total distance = 100 m + 400 m + 600 m
Total distance = 1100 m
Therefore, the total distance traveled by the car during this entire motion is
1100 m.
3
Question 4
Question
A car traveling at a constant velocity of 25 m/s passes a traffic light just as it
turns yellow. The driver knows that the light will turn red in 2.0 seconds. If the
acceleration of the car is -4.0 m/s2, what is the minimum distance the driver
must accelerate to avoid running the light?
Solution
Step 1: Let’s first determine the position of the car when the light turns red.
We can use the kinematic equation:
vf=vi+a·t
where vfis the final velocity, viis the initial velocity, ais the acceleration, and
tis the time.
Since the acceleration is constant, we can also use the equation:
x=xi+vi·t+1
2·a·t2
where xis the final position, xiis the initial position, viis the initial velocity,
ais the acceleration, and tis the time.
From the first equation, we know that the final velocity is 0 m/s when the
light turns red. So,
0 m/s = 25 m/s + (−4.0 m/s2)·2.0 s
Solving for x:
x= 0 + 25 ·2 + 1
2·(−4.0) ·(2)2
Step 2: Calculate the final position x.
x= 50 −8 = 42 m
Therefore, the minimum distance the driver must accelerate to avoid running
the light is 42 meters.
Question 5
Question
A car initially traveling at 20 m/s accelerates uniformly at 2 m/s2for 10 seconds.
What is the final velocity of the car?
4
Solution
Step 1: Identify the given variables and the unknown. Let vi= 20 m/s be the
initial velocity of the car, a= 2 m/s2be the acceleration, t= 10 s be the time,
and vfbe the final velocity.
Step 2: Use the kinematic equation to relate the variables: The kinematic
equation for uniformly accelerated motion in one dimension is:
vf=vi+a·t
Step 3: Plug in the given values to find the final velocity:
vf= 20 m/s + 2 m/s2·10 s
Step 4: Calculate the final velocity:
vf= 20 m/s + 20 m/s = 40 m/s
Therefore, the final velocity of the car is 40 m/s.
Question 6
Question
A car accelerates from rest with a constant acceleration of 3.5 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 2.5 m/s2until it comes to a
stop. What is the total distance covered by the car during this entire trip?
Solution
Step 1: Find the distance covered during the acceleration phase.
d1=1
2·a·t2=1
2·3.5 m/s2·(10 s)2= 175 m
Step 2: Find the distance covered during the constant speed phase.
d2=v·t= 3.5 m/s ·20 s = 70 m
Step 3: Find the distance covered during the deceleration phase.
d3=vi·t+1
2·a·t2= 3.5 m/s·20 s+1
2·(−2.5 m/s2)·(20 s)2= 70 m−250 m = −180 m
Step 4: Calculate the total distance covered by the car by summing the
distances obtained in each phase.
Total distance = |d1|+|d2|+|d3|= 175 m + 70 m + | − 180 m|= 425 m
Therefore, the total distance covered by the car during the entire trip is 425
meters.
5
Question 7
Question
A car accelerates uniformly from rest and reaches a speed of 22 m/s in 8 seconds.
Calculate the distance the car travels during this time.
Solution
Step 1: Determine the acceleration of the car using the equation of motion:
v=u+at
where v= 22 m/s (final velocity), u= 0 m/s (initial velocity, as the car starts
from rest), t= 8 s (time taken).
Substitute the values into the equation:
22 = 0 + a×8
a=22
8
a= 2.75 m/s2
Step 2: Use the equation of motion to find the distance traveled by the car:
s=ut +1
2at2
where sis the distance traveled by the car.
Substitute the known values into the equation:
s= 0 ×8 + 1
2×2.75 ×82
s= 0 + 0.5×2.75 ×64
s= 0 + 0.5×176
s= 0 + 88
s= 88 m
Therefore, the car travels a distance of 88 meters during the 8-second period.
Question 8
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 8.0
seconds. After this time, the car maintains a constant velocity. How far does
the car travel during the first 8.0 seconds of motion?
6
Solution
Step 1: Find the final velocity of the car after 8.0 seconds using the equation
v=u+at, where vis the final velocity, uis the initial velocity (0 m/s in this
case), ais the acceleration, and tis the time.
Final velocity: v= 0 + (3.0 m/s2×8.0 s) = 24.0 m/s
Step 2: Calculate the distance traveled during the first 8.0 seconds using the
equation s=ut +1
2at2, where sis the distance traveled.
Distance traveled: s= 0 ×8.0 + 1
2×3.0 m/s2×(8.0 s)2
s= 0 + 1
2×3.0 m/s2×64.0 s2
s=1
2×3.0×64.0 = 96.0 m
Therefore, the car travels a distance of 96.0 meters during the first 8.0 sec-
onds of motion.
Question 9
Question
A car accelerates uniformly from rest to a speed of 24 m/s in 6 seconds. How
far has the car traveled during this time interval?
Solution
Step 1: Identify the given variables.
The initial velocity of the car (u) is 0 m/s, the final velocity of the car (v) is 24
m/s, and the time interval (t) is 6 seconds.
Step 2: Use the kinematic equation to find the acceleration of the car.
The kinematic equation relating initial velocity (u), final velocity (v), accelera-
tion (a), and time interval (t) is:
v=u+at
Since the initial velocity (u) is 0, we have:
v=at
Solving for acceleration (a), we get:
a=v
t=24 m/s
6 s = 4 m/s2
7
Step 3: Use another kinematic equation to find the distance traveled by the
car.
The kinematic equation relating initial velocity (u), final velocity (v), accelera-
tion (a), and displacement (s) is:
v2=u2+ 2as
Since the initial velocity (u) is 0, the equation simplifies to:
v2= 2as
Solving for displacement (s), we get:
s=v2
2a=(24 m/s)2
2(4 m/s2)= 72 m
Therefore, the car has traveled 72 meters during this time interval.
Question 10
Question
A car is traveling on a straight road. It starts from rest and accelerates uniformly
at 2.5 m/s2for 8 seconds. Then it maintains a constant velocity for 10 seconds
before decelerating uniformly at 1.5 m/s2until it comes to a stop. What is the
total distance the car travels during this entire journey?
Solution
Step 1: Calculate the distance traveled during acceleration phase. The initial
velocity, u, is 0 m/s. The acceleration, a, is 2.5 m/s2. The time, t, is 8 seconds.
The distance traveled during acceleration can be calculated using the equa-
tion:
s=ut +1
2at2
Plugging in the values:
s= (0)(8) + 1
2(2.5)(8)2
s= 0 + 1
2(2.5)(64)
s= 0 + 80 = 80 meters
Therefore, during the acceleration phase, the car travels 80 meters.
Step 2: Calculate the distance traveled during constant velocity phase. The
velocity during this phase remains constant, so we can use the formula s=vt.
8
The velocity is the final velocity from the acceleration phase, which can be
calculated as:
v=u+at
v= 0 + 2.5(8)
v= 0 + 20 = 20 m/s
Thus, during the constant velocity phase, the car travels:
s=vt = (20)(10) = 200 meters
Step 3: Calculate the distance traveled during deceleration phase. The car
comes to a stop during the deceleration phase, so the final velocity is 0 m/s. The
deceleration, a, is 1.5 m/s2. The distance can be calculated using the formula:
v2=u2+ 2as
Since the final velocity is 0, the equation simplifies to:
0 = (20)2+ 2(−1.5)s
400 = −3s
s=−400
3=−133.3 meters
Since distance cannot be negative, we take the absolute value:
s= 133.3 meters
Step 4: Calculate the total distance traveled by the car during the entire
journey. The total distance is the sum of the distances traveled during each
phase:
Total distance = 80 + 200 + 133.3 = 413.3 meters
Therefore, the total distance the car travels during the entire journey is 413.3
meters.
Question 11
Question
A particle moves along a straight line according to the equation of motion x(t) =
4t3−3t2−2t+1, where xis in meters and tis in seconds. Determine the velocity
and acceleration of the particle when t= 2 s.
9
Solution
Step 1: To find the velocity of the particle, we differentiate the equation of
motion x(t) with respect to time tto get the velocity function v(t).
v(t) = dx
dt =d
dt(4t3−3t2−2t+ 1)
v(t) = 12t2−6t−2
Step 2: Substitute t= 2 s into the velocity function to find the velocity of
the particle at t= 2 s.
v(2) = 12(2)2−6(2) −2
v(2) = 48 −12 −2
v(2) = 34 m/s
Step 3: To find the acceleration of the particle, differentiate the velocity
function v(t) with respect to time tto get the acceleration function a(t).
a(t) = dv
dt =d
dt(12t2−6t−2)
a(t) = 24t−6
Step 4: Substitute t= 2 s into the acceleration function to find the acceler-
ation of the particle at t= 2 s.
a(2) = 24(2) −6
a(2) = 48 −6
a(2) = 42 m/s2
Therefore, when t= 2 s, the velocity of the particle is 34 m/s and the
acceleration of the particle is 42 m/s2.
Question 12
Question
A car accelerates from rest along a straight road. The car’s acceleration is given
by a(t)=4t−6 m/s2, where tis in seconds. Find the car’s velocity as a function
of time t.
10
Solution
Step 1: To find the car’s velocity as a function of time, we will integrate the
acceleration function a(t) with respect to time t.
Za(t)dt =Z(4t−6) dt
Step 2: Integrating each term separately, we get:
Za(t)dt =Z4t dt −Z6dt
Step 3: Integrating each term, we find:
v(t) = 2t2−6t+C
where v(t) represents the car’s velocity as a function of time tand Cis the
constant of integration. Step 4: To find the value of the constant C, we use the
initial condition that the car starts from rest, which means that v(0) = 0. Step
5: Substituting t= 0 and v(t) = 0 into the velocity function, we have:
0 = 2(0)2−6(0) + C
Step 6: Solving for C, we find C= 0. Step 7: Therefore, the car’s velocity as a
function of time is:
v(t)=2t2−6tm/s
Question 13
Question
A car is traveling along a straight road with a velocity of 20 m/s. The driver
applies the brakes, causing the car to decelerate at a rate of 4 m/s2. How far
does the car travel before coming to a stop?
Solution
Step 1: Identify the given variables. The initial velocity, v0, of the car is 20
m/s, the deceleration rate, a, is -4 m/s2(negative because it is in the opposite
direction of motion), and the final velocity, vf, is 0 m/s (since the car comes to
a stop).
Step 2: Use the kinematic equation with no time variable to solve for dis-
placement. The kinematic equation relates initial velocity, final velocity, accel-
eration, and displacement:
v2
f=v2
0+ 2a∆x
Step 3: Substitute the known values into the kinematic equation. Since
vf= 0, we have:
0 = (20)2+ 2(−4)∆x
11
Step 4: Solve for the displacement, ∆x.
0 = 400 −8∆x
8∆x= 400
∆x=400
8
∆x= 50 m
Therefore, the car travels 50 meters before coming to a stop.
Question 14
Question
An object is moving along a straight line with an initial velocity of 6 m/s. The
object undergoes acceleration according to the equation a= 2t, where ais the
acceleration in m/s2and tis the time in seconds. Find the object’s velocity as
a function of time and its position at t= 3 seconds.
Solution
Step 1: To find the velocity as a function of time, we need to integrate the
acceleration function with respect to time.
Za dt =Z2t dt
Zv dv =Z2t dt
Step 2: Integrating both sides gives us:
1
2v2=t2+C1
Where C1is the constant of integration.
Step 3: Given the initial velocity is 6 m/s at t= 0, we can find C1.
1
2×62= 0 + C1
C1= 18
Step 4: Substituting C1back into the equation:
1
2v2=t2+ 18
Step 5: To find the object’s velocity as a function of time, we solve for v:
v=p2t2+ 36
12
Step 6: Next, to find the object’s position at t= 3 seconds, we need to
integrate the velocity function with respect to time.
Zv dt =Zp2t2+ 36 dt
Step 7: Integrating both sides gives us the position function:
s=1
3(2t2+ 36)3/2+C2
Where C2is the constant of integration.
Step 8: Given the initial position is 0 at t= 0, we can find C2.
0 = 1
3(2 ×02+ 36)3/2+C2
C2=−36
Step 9: Substituting C2back into the equation:
s=1
3(2t2+ 36)3/2−36
Step 10: Substituting t= 3 into the position function gives us the object’s
position at t= 3 seconds.
s(3) = 1
3(2 ×32+ 36)3/2−36
s(3) = 1
3×543/2−36
s(3) = 54 −36
s(3) = 18
Therefore, the object’s velocity as a function of time is v=√2t2+ 36 and
its position at t= 3 seconds is 18 meters.
Question 15
Question
An object starts from rest and accelerates along a straight line. Its acceleration
as a function of time is given by a(t) = 2t−1 m/s2. Find the object’s velocity
as a function of time and its displacement after 3 seconds.
13
Solution
Step 1: To find the object’s velocity as a function of time, we can integrate the
acceleration function with respect to time:
Za(t)dt =Z(2t−1) dt
Step 2: Integrating the right side with respect to t, we get:
v(t) = Z(2t−1) dt =t2−t+C
where Cis the constant of integration.
Step 3: We know that the object starts from rest, therefore its initial velocity
is v(0) = 0. Substituting t= 0 and v= 0 into the velocity function gives us
C= 0. So the velocity as a function of time is:
v(t) = t2−tm/s
Step 4: To find the displacement after 3 seconds, we integrate the velocity
function over the time interval [0,3]:
Z3
0
v(t)dt =Z3
0
(t2−t)dt
Step 5: Integrating the right side with respect to t, we get:
Z3
0
v(t)dt =1
3t3−1
2t2
3
0
Step 6: Evaluating the integral at the limits of integration, we find the
displacement:
1
3(3)3−1
2(3)2−1
3(0)3−1
2(0)2=27
3−9
2
Step 7: Therefore, the displacement after 3 seconds is 27
3−9
2= 4.5 meters.
Question 16
Question
A car accelerates from rest along a straight road at 3.0 m/s2for 5.0 seconds. It
then maintains a constant velocity for 10 seconds before coming to a stop with
a constant deceleration of 2.0 m/s2. What is the total distance traveled by the
car during this entire motion?
14
Solution
Step 1: Find the distance traveled during acceleration. Given initial velocity
vi= 0 m/s, acceleration a= 3.0 m/s2, and time t= 5.0 s, we use the kinematic
equation:
d=vit+1
2at2
d= 0 + 1
2×3.0×(5.0)2
d= 0 + 22.5
d= 22.5 m
Step 2: Find the distance traveled at constant velocity. The velocity remains
constant for 10 seconds, so the distance traveled is:
d= velocity ×time = 3.0×10 = 30 m
Step 3: Find the distance traveled during deceleration. The car comes to
a stop with a constant deceleration of 2.0 m/s2. We need to find the distance
traveled during deceleration using the equation:
v2
f=v2
i+ 2ad
Since the final velocity vf= 0 m/s, the initial velocity vi= 3.0 m/s, the
deceleration a=−2.0 m/s2, we solve for d:
0 = (3.0)2+ 2 ×(−2.0) ×d
0=9−4d
4d= 9
d=9
4= 2.25 m
Step 4: Find the total distance traveled. The total distance traveled is the
sum of distances traveled during acceleration, constant velocity, and decelera-
tion:
dtotal = 22.5 + 30 + 2.25 = 54.75 m
Therefore, the total distance traveled by the car during this entire motion is
54.75 meters.
Question 17
Question
A car is initially at rest. It then accelerates along a straight road, such that
the velocity of the car at any time tis given by the function v(t) = 5t2−3t+ 2
where vis in m/s and tis in seconds. Determine the time tat which the car
comes to a stop.
15
Solution
Step 1: To find the time tat which the car comes to a stop, we need to determine
when the velocity v(t) is equal to zero.
Step 2: Set v(t) = 0:
5t2−3t+ 2 = 0
Step 3: Solve the quadratic equation by using the quadratic formula, t=
−b±√b2−4ac
2a, where a= 5, b=−3, and c= 2:
t=−(−3) ±p(−3)2−4·5·2
2·5
Step 4: Simplify the expression:
t=3±√9−40
10
Step 5: Since the discriminant is negative, there are no real solutions. This
means the car never comes to a stop along the road.
Question 18
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2for 8 seconds.
What is the final velocity of the car at the end of the 8-second interval?
Solution
Step 1: Identify the given variables.
Initial velocity vinitial = 0 m/s
Acceleration a= 2.5 m/s2
Time t= 8 s
Final velocity vfinal (unknown)
Step 2: Use the kinematic equation vfinal =vinitial +at to find the final
velocity.
Plugging in the given values, we get:
vfinal = 0 m/s + 2.5 m/s2×8 s
Step 3: Solve for the final velocity.
vfinal = 0 + 20 = 20 m/s
Answer: The final velocity of the car at the end of the 8-second interval is
20 m/s.
16
Question 19
Question
A car is traveling at a constant velocity of 25 m/s. At a certain instant, the
driver applies the brakes and comes to a stop after 50 m. If the deceleration of
the car is constant, what is the time it takes for the car to come to a stop?
Solution
Step 1: Identify the given variables and their values. The initial velocity of the
car, u= 25 m/s (since the car is traveling at a constant velocity). The final
velocity of the car, v= 0 (since the car comes to a stop). The displacement of
the car, s= 50 m. The acceleration of the car is given as the deceleration a,
which is opposite in direction to the motion of the car.
Step 2: Choose the appropriate kinematic equation to solve the problem.
The kinematic equation that relates initial velocity, final velocity, acceleration,
and displacement is:
v2=u2+ 2as
Step 3: Substitute the known values into the kinematic equation. Since the
final velocity is 0, the equation becomes:
02= (25 m/s)2+ 2a(50 m)
Step 4: Solve for the acceleration a.
0 = 625 + 100a
100a=−625
a=−6.25 m/s2
Step 5: Use the kinematic equation to find the time taken for the car to
stop. The kinematic equation relating initial velocity, acceleration, and time is:
v=u+at
Substitute the known values into the equation:
0 = 25 + (−6.25)t
6.25t= 25
t= 4 s
Therefore, it takes 4 seconds for the car to come to a stop.
17
Question 20
Question
A car accelerates uniformly from rest at 3.0 m/s
²
for 10 seconds, then maintains
a constant velocity for 20 seconds, and finally decelerates uniformly to a stop
in 5 seconds. Calculate the total distance traveled by the car during this entire
motion.
Solution
Step 1: Find the distance traveled during acceleration phase.
Given: Initial velocity, u= 0 m/s (car starts from rest) Acceleration, a= 3.0
m/s
²
Time, t= 10 s
Using the equation of motion: s=ut +1
2at2
Substitute the known values: s1= 0 ×10 + 1
2×3.0×(10)2s1= 0 + 0.5×
3.0×100 s1= 0 + 0.5×300 s1= 0 + 150 s1= 150 m
Therefore, the distance traveled during the acceleration phase is 150 m.
Step 2: Find the distance traveled during constant velocity phase.
Given: Velocity during constant velocity phase, v=u+at Constant velocity,
which means a= 0 Time, t= 20 s
Using the equation: s=vt
Substitute the known values: s2= 3.0×20 s2= 60 m
Therefore, the distance traveled during the constant velocity phase is 60 m.
Step 3: Find the distance traveled during deceleration phase.
Given: Final velocity, v= 0 m/s (car comes to a stop) Deceleration, which
is negative acceleration, a=−3.0 m/s
²
Time, t= 5 s
Using the equation of motion: s=ut +1
2at2
Substitute the known values: s3= 3.0×5 + 1
2×−3.0×(5)2s3= 15 −0.5×
3.0×25 s3= 15 −0.5×75 s3= 15 −37.5s3=−22.5 m
The negative sign indicates the direction of motion is opposite to the positive
direction.
Step 4: Calculate the total distance traveled by summing up the distances.
Total distance traveled, stotal =s1+s2+s3stotal = 150 + 60 −22.5stotal =
187.5 m
Therefore, the total distance traveled by the car during the entire motion is
187.5 m.
Question 21
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
It then maintains a constant velocity for 20 seconds before decelerating at a rate
of −3 m/s2until it comes to a stop. Find the total distance the car travels during
this entire process.
18
Solution
Step 1: Find the distance traveled during acceleration. The distance traveled
during acceleration can be found using the equation:
d=1
2·a·t2
where ais the acceleration and tis the time. Substitute a= 2 m/s2and t= 10 s:
d=1
2·2 m/s2·(10 s)2= 100 m
Step 2: Find the distance traveled during constant velocity. The distance
traveled during constant velocity can be found using the equation:
d=v·t
where vis the constant velocity and tis the time. The car maintains a constant
velocity for 20 seconds, so:
d=v·20
Since the car maintains a constant velocity during this time, the distance trav-
eled is simply the velocity multiplied by time. However, as the car starts from
rest and accelerates to this velocity, the distance traveled chronologically is the
sum of distances traveled during acceleration, constant velocity, and decelera-
tion.
Step 3: Find the distance traveled during deceleration. The distance traveled
during deceleration can also be found using the equation:
d=1
2·a·t2
Substitute a=−3 m/s2and the car comes to a stop, so the final velocity is 0.
0 = vf+a·t
t=−vf
a=0
−3= 0 s
So, the car comes to a stop immediately after moving at constant velocity for
20 seconds. The distance traveled during deceleration is:
d=1
2·(−3 m/s2)·(0 s)2= 0
Step 4: Calculate the total distance. The total distance traveled is the sum
of the distances traveled during acceleration, constant velocity, and deceleration:
Total distance = 100 m + v·20 + 0 = 100 m + v·20
Since the velocity is constant during the 20 seconds after acceleration, v=
2 m/s ·10 s = 20 m/s.
Total distance = 100 m + 20 m/s ·20 s = 100 m + 400 m = 500 m
Therefore, the total distance the car travels during this entire process is 500
meters.
19
Question 22
Question
A particle moves in a straight line with a velocity described by the equation
v(t) = 8 −4t+ 2t2, where vis the velocity in m/s and tis the time in seconds.
Determine the particle’s acceleration as a function of time.
Solution
Step 1: To find the particle’s acceleration as a function of time, we differentiate
the velocity function with respect to time. The acceleration a(t) is given by the
derivative of the velocity function v(t).
Step 1: a(t) = dv
dt
Step 2: Differentiate v(t) term by term:
Step 2: a(t) = d
dt(8 −4t+ 2t2)
Step 3: The derivative of a constant is 0, and the differentiation of −4tand
2t2with respect to tis −4 and 4t, respectively.
Step 3: a(t) = −4+4t
Step 4: Therefore, the particle’s acceleration as a function of time is a(t) =
−4 + 4t. This means that the acceleration is changing linearly with time, with
a constant acceleration of 4 m/s2.
Step 4: a(t) = −4+4t
Question 23
Question
An object is thrown straight up into the air from the ground with an initial
velocity of 20 m/s. How high above the ground does the object travel before
coming back down?
Solution
Step 1: Define the upward motion as positive.
Step 2: Identify the given quantities: Initial velocity, v0= 20 m/s
Acceleration due to gravity, a=−9.81 m/s2(negative because it acts downward)
Step 3: Use the kinematic equation for displacement to find the height the
object travels:
v2=v2
0+ 2a∆y
20
Substitute the given values:
0 = (20 m/s)2+ 2(−9.81 m/s2)∆y
Step 4: Solve for ∆y:
0 = 400 −19.62∆y
19.62∆y= 400
∆y=400
19.62
∆y≈20.38 m
Therefore, the object travels approximately 20.38 meters above the ground
before coming back down.
Question 24
Question
An object moves along a straight line according to the equation of motion x(t) =
5t3−3t2+2, where xis in meters and tis in seconds. Find the maximum velocity
of the object and the time at which it occurs.
Solution
Step 1: Find the velocity function by taking the derivative of the position
function with respect to time.
Step 1: v(t) = dx
dt =d
dt(5t3−3t2+ 2)
Step 2: Differentiate each term separately.
Step 2: v(t) = 15t2−6t
Step 3: To find the maximum velocity, we need to find the time when the
velocity is maximum. This occurs when the acceleration is zero. So, differentiate
the velocity function.
Step 3: a(t) = dv
dt =d
dt(15t2−6t)
Step 4: Differentiate each term separately.
Step 4: a(t) = 30t−6
Step 5: Set the acceleration function equal to zero and solve for t.
Step 5: 0 = 30t−6
21
30t= 6
t=6
30 = 0.2 s
Step 6: Find the maximum velocity by substituting t= 0.2 into the velocity
function.
Step 6: v(0.2) = 15(0.2)2−6(0.2)
v(0.2) = 15(0.04) −1.2
v(0.2) = 0.6−1.2 = −0.6 m/s
Therefore, the maximum velocity of the object is −0.6 m/s and it occurs at
t= 0.2 s.
Question 25
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 5.0
seconds. It then maintains a constant velocity for 10 seconds before coming
to a stop with a constant deceleration of 2.0 m/s2. What is the total distance
traveled by the car during this entire motion?
Solution
Step 1: Calculate the distance covered during acceleration. The distance covered
during acceleration can be calculated using the equation:
d=1
2at2
where ais the acceleration and tis the time taken. Plugging in the values, we
get:
d=1
2×3.0 m/s2×(5.0 s)2
d=1
2×3.0×25
d=75
2
d= 37.5 m
Step 2: Calculate the distance covered during constant velocity. The distance
covered during constant velocity can be calculated using the equation:
d=vt
where vis the constant velocity and tis the time taken. Given that the car
maintains a constant velocity, the distance covered is:
d=v×10
22
Since the car maintained a constant velocity, the distance covered is:
d= 0 ×10 = 0 m
Step 3: Calculate the distance covered during deceleration. The distance
covered during deceleration can be calculated using the equation:
d=1
2at2
where ais the deceleration and tis the time taken. Plugging in the values, we
get:
d=1
2×2.0 m/s2×(10 s)2
d=1
2×2.0× ×100
d= 100 m
Step 4: Calculate the total distance traveled. The total distance traveled
is the sum of the distances covered during acceleration, constant velocity, and
deceleration:
Total distance = 37.5 m + 0 m + 100 m
Total distance = 137.5 m
Therefore, the total distance traveled by the car during this entire motion is
137.5 meters.
Question 26
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2. Find the acceler-
ation of the car when the velocity is 24 m/s.
Solution
Step 1: Identify known values
The initial velocity of the car, u, is 0 m/s.
The final velocity of the car, v, is 24 m/s.
The acceleration of the car, a, is 2.0 m/s2.
Step 2: Choose the appropriate equation of motion
We can use the equation relating final velocity, initial velocity, acceleration, and
displacement:
v2=u2+ 2as
Step 3: Substitute known values into the equation
Substitute the known values into the equation:
23
(24 m/s)2= (0 m/s)2+ 2(2.0 m/s2)·s
Step 4: Solve for displacement (s)
576 m2/s2= 4.0 m/s2·s
s=576 m2/s2
4.0 m/s2= 144 m
Step 5: Calculate the acceleration at v= 24 m/s
Now that we have found the displacement (s), we can use the following equation
to find the acceleration (a) at v= 24 m/s:
v2=u2+ 2as
(24 m/s)2= (0 m/s)2+ 2a(144 m)
576 m2/s2= 288am
a=576 m2/s2
288 m = 2.0 m/s2
Therefore, the acceleration of the car when the velocity is 24 m/s is 2.0 m/s2.
Question 27
Question
A car starting from rest accelerates in a straight line at a constant rate of 2
m/s2. Calculate the time it takes for the car to reach a speed of 20 m/s.
Solution
Step 1: Identify the given values.
The initial velocity uof the car is 0 m/s, the acceleration ais 2 m/s2, and the
final velocity vis 20 m/s.
Step 2: Use the kinematic equation v=u+at
Substitute the known values into the equation:
v=u+at
20 = 0 + 2t
Step 3: Solve for time t.
20 = 2t
24
t=20
2
t= 10 seconds
Answer: It takes the car 10 seconds to reach a speed of 20 m/s.
Question 28
Question
A car traveling initially at a speed of 20 m/s accelerates uniformly at 2 m/s2for
10 seconds, then maintains a constant speed for the next 20 seconds. Calculate
the total distance the car travels during this time period.
Solution
Step 1: Calculate the distance covered during acceleration period. Given: Initial
velocity (u) = 20 m/s, acceleration (a) = 2 m/s2, time of acceleration (t) = 10
seconds.
Using the kinematic equation:
s=ut +1
2at2
We can substitute the given values to find the distance covered during ac-
celeration.
s= (20 m/s)(10 s) + 1
2(2 m/s2)(10 s)2
s= 200 m + 1
2(2)(100)
s= 200 m + 100
s= 300 m
Therefore, the distance covered during acceleration is 300 m.
Step 2: Calculate the distance covered during constant speed. During the
constant speed period, the car maintains a speed of 20 m/s for 20 seconds. The
distance covered during this time period is:
s= speed ×time = 20 m/s ×20 s = 400 m
Step 3: Calculate the total distance traveled. The total distance traveled is
the sum of the distances covered during acceleration and constant speed.
Total distance = 300 m + 400 m = 700 m
Therefore, the total distance the car travels during this time period is 700
meters.
25
Question 29
Question
A car starts from rest and accelerates along a straight road at 3.0 m/s2. At the
same instant, a truck passes the car traveling at a constant speed of 30 m/s.
How far down the road will the car overtake the truck?
Solution
Step 1: First, we find the time it takes for the car to overtake the truck. Let t
be the time it takes for the car to overtake the truck.
Step 2: We can set up equations for the motion of the car and the truck.
For the car, the position at time tis given by:
xcar =1
2at2
Step 3: For the truck, the position at time tis given by:
xtruck =vt
Step 4: When the car overtakes the truck, the positions are equal:
1
2at2=vt
Step 5: Substituting the given values for aand vinto the equation, we get:
1
2(3.0 m/s2)t2= (30 m/s)t
Step 6: Further simplifying the equation, we have:
1.5t2= 30t
Step 7: Dividing by t(assuming t= 0), we get:
1.5t= 30
t=30
1.5
t= 20 s
Step 8: Finally, we can find the distance the car travels to overtake the truck
by substituting tback into the equation for the car’s position:
xcar =1
2(3.0 m/s2)(20 s)2
xcar =1
2(3.0)(400) m
xcar = 600 m
Therefore, the car will overtake the truck after traveling 600 meters down
the road.
26
Question 30
Question
A car accelerates from rest at a constant rate of 3.0 m/s2. How far does the car
travel in the first 5.0 seconds?
Solution
Step 1: We will first find the final velocity of the car after 5.0 seconds using the
kinematic equation:
vf=vi+at
where vfis the final velocity, viis the initial velocity (in this case, 0 m/s), ais
the acceleration, and tis the time. Substituting the values:
vf= 0 + (3.0 m/s2)(5.0 s)
vf= 15.0 m/s
Step 2: Next, we will find the distance traveled by the car using the equation:
d=vit+1
2at2
where dis the distance traveled, viis the initial velocity, tis the time, and ais
the acceleration. Substituting the values:
d= 0 + 1
2(3.0 m/s2)(5.0 s)2
d=1
2(3.0 m/s2)(25 s2)
d=1
2(75 m)
d= 37.5 m
Therefore, the car travels a distance of 37.5 meters in the first 5.0 seconds.
Question 31
Question
A car initially at rest starts moving along a straight road. The car accelerates
uniformly over a distance of 200 meters and reaches a final velocity of 20 m/s.
Calculate the time it takes for the car to reach this final velocity.
27
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity (u) = 0 m/s (car is initially at rest)
Final velocity (v) = 20 m/s
Distance traveled (s) = 200 m
Acceleration (a) = ?
Time taken (t) = ?
Step 2: Use the kinematic equation relating displacement, initial velocity,
final velocity, acceleration, and time:
v=u+at
Step 3: Rearrange the equation to solve for acceleration:
a=v−u
t
Step 4: Since the car starts from rest, u= 0. Substitute the known values:
a=20 m/s −0 m/s
t
a=20 m/s
t
Step 5: Use the kinematic equation relating displacement, initial velocity,
acceleration, and time:
s=ut +1
2at2
Step 6: Substitute the known values:
200 m = 0 m/s ×t+1
2at2
200 = 1
2at2
400 = at2
Step 7: Substitute the expression for acceleration found in Step 4 into the
equation in Step 6:
400 = 20 m/s
t·t2
400 = 20t
t= 20 s
Therefore, it takes the car 20 seconds to reach a final velocity of 20 m/s.
28
Question 32
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 10
seconds. What is the total distance traveled by the car during this time interval?
Solution
Step 1: Determine the final velocity of the car using the equation v=u+at,
where vis the final velocity, uis the initial velocity (0 m/s in this case), ais
the acceleration, and tis the time.
Final velocity = 0 + (3.0 m/s2)(10 s)
Final velocity = 30 m/s
Step 2: Calculate the total distance traveled by the car using the equation
for distance covered under constant acceleration, s=ut +1
2at2.
s= (0 m/s)(10 s) + 1
2(3.0 m/s2)(10 s)2
s= 0 + 0.5(3.0)(100)
s= 150 m
Therefore, the total distance traveled by the car during this time interval is
150 meters.
Question 33
Question
A car starts from rest and accelerates uniformly at 2.5 m/s2along a straight
track. How far has the car traveled when it reaches a speed of 25 m/s?
Solution
Step 1: Let’s denote the initial velocity of the car as vi= 0 m/s, the acceleration
as a= 2.5 m/s2, the final velocity as vf= 25 m/s, and the distance traveled as
d. We can use the following kinematic equation to relate these variables:
v2
f=v2
i+ 2ad
Step 2: Substituting the given values into the equation, we get:
(25 m/s)2= (0 m/s)2+ 2(2.5 m/s2)·d
Step 3: Simplifying the equation, we have:
29
625 m/s2= 5 m/s2·d
Step 4: Solving for d, we find:
d=625 m/s2
5 m/s2= 125 m
So, the car has traveled 125 meters when it reaches a speed of 25 m/s.
Question 34
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the car decelerates at a constant rate of 4 m/s2until it comes
to a stop. Determine the total distance traveled by the car during this time
period.
Solution
Step 1: Let’s first determine the distance traveled during the acceleration phase.
The velocity of the car after 10 seconds can be found using the equation of
motion:
v=u+at,
where uis the initial velocity (0 m/s), ais the acceleration (2 m/s2), and tis
the time (10 seconds).
v= 0 + 2 ×10 = 20 m/s.
The distance traveled during the acceleration phase can be found using:
s=ut +1
2at2,
where sis the distance traveled.
s= 0 ×10 + 1
2×2×102=1
2×2×100 = 100 m.
Step 2: Next, let’s determine the distance traveled during the deceleration
phase. To find the time taken to stop during deceleration, we use:
v=u+at,
where uis the initial velocity (20 m/s), ais the deceleration (-4 m/s2), and vis
the final velocity (0 m/s).
0 = 20 −4t⇒t=20
4= 5 s.
30
The distance traveled during deceleration can be found using:
s=ut +1
2at2,
where uis the initial velocity (20 m/s), tis the time (5 seconds), and ais the
deceleration (-4 m/s2).
s= 20 ×5 + 1
2× −4×52= 100 −50 = 50 m.
Step 3: Finally, the total distance traveled by the car is the sum of the
distances traveled during acceleration and deceleration.
Total distance = 100 m + 50 m = 150 m.
Therefore, the total distance traveled by the car during this time period is
150 meters.
Question 35
Question
A stone is thrown straight upward with an initial velocity of 20 m/s from the
edge of a cliff 100 m high. 1. How long will it take for the stone to hit the
ground at the foot of the cliff? 2. With what speed does it strike the ground?
Solution
Let’s first find the time it takes for the stone to hit the ground. We can use the
kinematic equation for vertical motion:
y=vit+1
2at2
where: y=−100 m (negative because down is taken as the negative direction),
vi= 20 m/s, a=−9.8 m/s2(acceleration due to gravity), and we want to find
t.
Step 1: Find the time it takes for the stone to hit the ground. Plugging in
the values, we have:
−100 = 20t−1
2·9.8·t2
Simplifying gives us:
4.9t2−20t−100 = 0
This is a quadratic equation where:
a= 4.9, b =−20, c =−100
31
Solution
Step 1: Identify the given variables and the unknown variable. Let a= 2.0 m/s2
be the acceleration of the car, vf= 25 m/s be the final velocity, and tbe the
time taken for the car to reach a speed of 25 m/s.
Step 2: Find the equation connecting the variables. We know that the final
velocity of an object accelerated from rest is given by the equation:
vf=at
Step 3: Substitute the known values into the equation. Substitute vf= 25
m/s and a= 2.0 m/s2into the equation:
25 = 2.0t
Step 4: Solve for the unknown variable. Solving for t, we get:
t=25
2.0= 12.5 s
Therefore, it takes 12.5 seconds for the car to reach a speed of 25 m/s.
Question 3
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the car maintains a constant velocity for 20 seconds before
coming to a stop with a constant deceleration of 3 m/s2. Calculate the total
distance traveled by the car during this entire motion.
Solution
Step 1: Find the distance traveled during acceleration phase. The distance
covered during acceleration can be calculated using the equation:
s=ut +1
2at2
where uis the initial velocity, ais the acceleration, tis the time.
Given that the initial velocity, u= 0 m/s, acceleration, a= 2 m/s2, and
time, t= 10 s, we can plug these values into the equation to find the distance
traveled during acceleration:
s= 0 ×10 + 1
2×2×(10)2
s= 0 + 1
2×2×100
2
s= 0 + 1 ×100
s= 100 m
Therefore, the distance covered during acceleration phase is 100 m.
Step 2: Find the distance traveled during constant velocity phase. Since the
car maintains a constant velocity during this phase, the distance traveled can
be calculated using the equation:
s=vt
where vis the constant velocity and tis the time.
Given that the constant velocity, v= 2 ×10 = 20 m/s, and time, t= 20 s,
we can find the distance traveled during constant velocity motion:
s= 20 ×20
s= 400 m
Therefore, the distance covered during the constant velocity phase is 400 m.
Step 3: Find the distance traveled during deceleration phase. The distance
covered during deceleration can be calculated using the same equation as for
acceleration:
s=vt +1
2at2
where vis the final velocity, ais the deceleration, tis the time.
Since the car comes to a stop at the end of the deceleration phase, the final
velocity, v= 0 m/s, deceleration, a= 3 m/s2, and time, t= 20 s, we can plug
these values into the equation to find the distance traveled during deceleration:
s= 0 ×20 + 1
2×3×(20)2
s= 0 + 1
2×3×400
s= 0 + 1.5×400
s= 600 m
Therefore, the distance covered during the deceleration phase is 600 m.
Step 4: Calculate the total distance traveled by the car. The total distance
traveled by the car is the sum of distances traveled during acceleration, constant
velocity, and deceleration phases: Total distance = 100 m + 400 m + 600 m
Total distance = 1100 m
Therefore, the total distance traveled by the car during this entire motion is
1100 m.
3
Question 4
Question
A car traveling at a constant velocity of 25 m/s passes a traffic light just as it
turns yellow. The driver knows that the light will turn red in 2.0 seconds. If the
acceleration of the car is -4.0 m/s2, what is the minimum distance the driver
must accelerate to avoid running the light?
Solution
Step 1: Let’s first determine the position of the car when the light turns red.
We can use the kinematic equation:
vf=vi+a·t
where vfis the final velocity, viis the initial velocity, ais the acceleration, and
tis the time.
Since the acceleration is constant, we can also use the equation:
x=xi+vi·t+1
2·a·t2
where xis the final position, xiis the initial position, viis the initial velocity,
ais the acceleration, and tis the time.
From the first equation, we know that the final velocity is 0 m/s when the
light turns red. So,
0 m/s = 25 m/s + (−4.0 m/s2)·2.0 s
Solving for x:
x= 0 + 25 ·2 + 1
2·(−4.0) ·(2)2
Step 2: Calculate the final position x.
x= 50 −8 = 42 m
Therefore, the minimum distance the driver must accelerate to avoid running
the light is 42 meters.
Question 5
Question
A car initially traveling at 20 m/s accelerates uniformly at 2 m/s2for 10 seconds.
What is the final velocity of the car?
4
Solution
Step 1: Identify the given variables and the unknown. Let vi= 20 m/s be the
initial velocity of the car, a= 2 m/s2be the acceleration, t= 10 s be the time,
and vfbe the final velocity.
Step 2: Use the kinematic equation to relate the variables: The kinematic
equation for uniformly accelerated motion in one dimension is:
vf=vi+a·t
Step 3: Plug in the given values to find the final velocity:
vf= 20 m/s + 2 m/s2·10 s
Step 4: Calculate the final velocity:
vf= 20 m/s + 20 m/s = 40 m/s
Therefore, the final velocity of the car is 40 m/s.
Question 6
Question
A car accelerates from rest with a constant acceleration of 3.5 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 2.5 m/s2until it comes to a
stop. What is the total distance covered by the car during this entire trip?
Solution
Step 1: Find the distance covered during the acceleration phase.
d1=1
2·a·t2=1
2·3.5 m/s2·(10 s)2= 175 m
Step 2: Find the distance covered during the constant speed phase.
d2=v·t= 3.5 m/s ·20 s = 70 m
Step 3: Find the distance covered during the deceleration phase.
d3=vi·t+1
2·a·t2= 3.5 m/s·20 s+1
2·(−2.5 m/s2)·(20 s)2= 70 m−250 m = −180 m
Step 4: Calculate the total distance covered by the car by summing the
distances obtained in each phase.
Total distance = |d1|+|d2|+|d3|= 175 m + 70 m + | − 180 m|= 425 m
Therefore, the total distance covered by the car during the entire trip is 425
meters.
5
Question 7
Question
A car accelerates uniformly from rest and reaches a speed of 22 m/s in 8 seconds.
Calculate the distance the car travels during this time.
Solution
Step 1: Determine the acceleration of the car using the equation of motion:
v=u+at
where v= 22 m/s (final velocity), u= 0 m/s (initial velocity, as the car starts
from rest), t= 8 s (time taken).
Substitute the values into the equation:
22 = 0 + a×8
a=22
8
a= 2.75 m/s2
Step 2: Use the equation of motion to find the distance traveled by the car:
s=ut +1
2at2
where sis the distance traveled by the car.
Substitute the known values into the equation:
s= 0 ×8 + 1
2×2.75 ×82
s= 0 + 0.5×2.75 ×64
s= 0 + 0.5×176
s= 0 + 88
s= 88 m
Therefore, the car travels a distance of 88 meters during the 8-second period.
Question 8
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 8.0
seconds. After this time, the car maintains a constant velocity. How far does
the car travel during the first 8.0 seconds of motion?
6
Solution
Step 1: Find the final velocity of the car after 8.0 seconds using the equation
v=u+at, where vis the final velocity, uis the initial velocity (0 m/s in this
case), ais the acceleration, and tis the time.
Final velocity: v= 0 + (3.0 m/s2×8.0 s) = 24.0 m/s
Step 2: Calculate the distance traveled during the first 8.0 seconds using the
equation s=ut +1
2at2, where sis the distance traveled.
Distance traveled: s= 0 ×8.0 + 1
2×3.0 m/s2×(8.0 s)2
s= 0 + 1
2×3.0 m/s2×64.0 s2
s=1
2×3.0×64.0 = 96.0 m
Therefore, the car travels a distance of 96.0 meters during the first 8.0 sec-
onds of motion.
Question 9
Question
A car accelerates uniformly from rest to a speed of 24 m/s in 6 seconds. How
far has the car traveled during this time interval?
Solution
Step 1: Identify the given variables.
The initial velocity of the car (u) is 0 m/s, the final velocity of the car (v) is 24
m/s, and the time interval (t) is 6 seconds.
Step 2: Use the kinematic equation to find the acceleration of the car.
The kinematic equation relating initial velocity (u), final velocity (v), accelera-
tion (a), and time interval (t) is:
v=u+at
Since the initial velocity (u) is 0, we have:
v=at
Solving for acceleration (a), we get:
a=v
t=24 m/s
6 s = 4 m/s2
7
Step 3: Use another kinematic equation to find the distance traveled by the
car.
The kinematic equation relating initial velocity (u), final velocity (v), accelera-
tion (a), and displacement (s) is:
v2=u2+ 2as
Since the initial velocity (u) is 0, the equation simplifies to:
v2= 2as
Solving for displacement (s), we get:
s=v2
2a=(24 m/s)2
2(4 m/s2)= 72 m
Therefore, the car has traveled 72 meters during this time interval.
Question 10
Question
A car is traveling on a straight road. It starts from rest and accelerates uniformly
at 2.5 m/s2for 8 seconds. Then it maintains a constant velocity for 10 seconds
before decelerating uniformly at 1.5 m/s2until it comes to a stop. What is the
total distance the car travels during this entire journey?
Solution
Step 1: Calculate the distance traveled during acceleration phase. The initial
velocity, u, is 0 m/s. The acceleration, a, is 2.5 m/s2. The time, t, is 8 seconds.
The distance traveled during acceleration can be calculated using the equa-
tion:
s=ut +1
2at2
Plugging in the values:
s= (0)(8) + 1
2(2.5)(8)2
s= 0 + 1
2(2.5)(64)
s= 0 + 80 = 80 meters
Therefore, during the acceleration phase, the car travels 80 meters.
Step 2: Calculate the distance traveled during constant velocity phase. The
velocity during this phase remains constant, so we can use the formula s=vt.
8
The velocity is the final velocity from the acceleration phase, which can be
calculated as:
v=u+at
v= 0 + 2.5(8)
v= 0 + 20 = 20 m/s
Thus, during the constant velocity phase, the car travels:
s=vt = (20)(10) = 200 meters
Step 3: Calculate the distance traveled during deceleration phase. The car
comes to a stop during the deceleration phase, so the final velocity is 0 m/s. The
deceleration, a, is 1.5 m/s2. The distance can be calculated using the formula:
v2=u2+ 2as
Since the final velocity is 0, the equation simplifies to:
0 = (20)2+ 2(−1.5)s
400 = −3s
s=−400
3=−133.3 meters
Since distance cannot be negative, we take the absolute value:
s= 133.3 meters
Step 4: Calculate the total distance traveled by the car during the entire
journey. The total distance is the sum of the distances traveled during each
phase:
Total distance = 80 + 200 + 133.3 = 413.3 meters
Therefore, the total distance the car travels during the entire journey is 413.3
meters.
Question 11
Question
A particle moves along a straight line according to the equation of motion x(t) =
4t3−3t2−2t+1, where xis in meters and tis in seconds. Determine the velocity
and acceleration of the particle when t= 2 s.
9
Solution
Step 1: To find the velocity of the particle, we differentiate the equation of
motion x(t) with respect to time tto get the velocity function v(t).
v(t) = dx
dt =d
dt(4t3−3t2−2t+ 1)
v(t) = 12t2−6t−2
Step 2: Substitute t= 2 s into the velocity function to find the velocity of
the particle at t= 2 s.
v(2) = 12(2)2−6(2) −2
v(2) = 48 −12 −2
v(2) = 34 m/s
Step 3: To find the acceleration of the particle, differentiate the velocity
function v(t) with respect to time tto get the acceleration function a(t).
a(t) = dv
dt =d
dt(12t2−6t−2)
a(t) = 24t−6
Step 4: Substitute t= 2 s into the acceleration function to find the acceler-
ation of the particle at t= 2 s.
a(2) = 24(2) −6
a(2) = 48 −6
a(2) = 42 m/s2
Therefore, when t= 2 s, the velocity of the particle is 34 m/s and the
acceleration of the particle is 42 m/s2.
Question 12
Question
A car accelerates from rest along a straight road. The car’s acceleration is given
by a(t)=4t−6 m/s2, where tis in seconds. Find the car’s velocity as a function
of time t.
10
Solution
Step 1: To find the car’s velocity as a function of time, we will integrate the
acceleration function a(t) with respect to time t.
Za(t)dt =Z(4t−6) dt
Step 2: Integrating each term separately, we get:
Za(t)dt =Z4t dt −Z6dt
Step 3: Integrating each term, we find:
v(t) = 2t2−6t+C
where v(t) represents the car’s velocity as a function of time tand Cis the
constant of integration. Step 4: To find the value of the constant C, we use the
initial condition that the car starts from rest, which means that v(0) = 0. Step
5: Substituting t= 0 and v(t) = 0 into the velocity function, we have:
0 = 2(0)2−6(0) + C
Step 6: Solving for C, we find C= 0. Step 7: Therefore, the car’s velocity as a
function of time is:
v(t)=2t2−6tm/s
Question 13
Question
A car is traveling along a straight road with a velocity of 20 m/s. The driver
applies the brakes, causing the car to decelerate at a rate of 4 m/s2. How far
does the car travel before coming to a stop?
Solution
Step 1: Identify the given variables. The initial velocity, v0, of the car is 20
m/s, the deceleration rate, a, is -4 m/s2(negative because it is in the opposite
direction of motion), and the final velocity, vf, is 0 m/s (since the car comes to
a stop).
Step 2: Use the kinematic equation with no time variable to solve for dis-
placement. The kinematic equation relates initial velocity, final velocity, accel-
eration, and displacement:
v2
f=v2
0+ 2a∆x
Step 3: Substitute the known values into the kinematic equation. Since
vf= 0, we have:
0 = (20)2+ 2(−4)∆x
11
Step 4: Solve for the displacement, ∆x.
0 = 400 −8∆x
8∆x= 400
∆x=400
8
∆x= 50 m
Therefore, the car travels 50 meters before coming to a stop.
Question 14
Question
An object is moving along a straight line with an initial velocity of 6 m/s. The
object undergoes acceleration according to the equation a= 2t, where ais the
acceleration in m/s2and tis the time in seconds. Find the object’s velocity as
a function of time and its position at t= 3 seconds.
Solution
Step 1: To find the velocity as a function of time, we need to integrate the
acceleration function with respect to time.
Za dt =Z2t dt
Zv dv =Z2t dt
Step 2: Integrating both sides gives us:
1
2v2=t2+C1
Where C1is the constant of integration.
Step 3: Given the initial velocity is 6 m/s at t= 0, we can find C1.
1
2×62= 0 + C1
C1= 18
Step 4: Substituting C1back into the equation:
1
2v2=t2+ 18
Step 5: To find the object’s velocity as a function of time, we solve for v:
v=p2t2+ 36
12
Step 6: Next, to find the object’s position at t= 3 seconds, we need to
integrate the velocity function with respect to time.
Zv dt =Zp2t2+ 36 dt
Step 7: Integrating both sides gives us the position function:
s=1
3(2t2+ 36)3/2+C2
Where C2is the constant of integration.
Step 8: Given the initial position is 0 at t= 0, we can find C2.
0 = 1
3(2 ×02+ 36)3/2+C2
C2=−36
Step 9: Substituting C2back into the equation:
s=1
3(2t2+ 36)3/2−36
Step 10: Substituting t= 3 into the position function gives us the object’s
position at t= 3 seconds.
s(3) = 1
3(2 ×32+ 36)3/2−36
s(3) = 1
3×543/2−36
s(3) = 54 −36
s(3) = 18
Therefore, the object’s velocity as a function of time is v=√2t2+ 36 and
its position at t= 3 seconds is 18 meters.
Question 15
Question
An object starts from rest and accelerates along a straight line. Its acceleration
as a function of time is given by a(t) = 2t−1 m/s2. Find the object’s velocity
as a function of time and its displacement after 3 seconds.
13
Solution
Step 1: To find the object’s velocity as a function of time, we can integrate the
acceleration function with respect to time:
Za(t)dt =Z(2t−1) dt
Step 2: Integrating the right side with respect to t, we get:
v(t) = Z(2t−1) dt =t2−t+C
where Cis the constant of integration.
Step 3: We know that the object starts from rest, therefore its initial velocity
is v(0) = 0. Substituting t= 0 and v= 0 into the velocity function gives us
C= 0. So the velocity as a function of time is:
v(t) = t2−tm/s
Step 4: To find the displacement after 3 seconds, we integrate the velocity
function over the time interval [0,3]:
Z3
0
v(t)dt =Z3
0
(t2−t)dt
Step 5: Integrating the right side with respect to t, we get:
Z3
0
v(t)dt =1
3t3−1
2t2
3
0
Step 6: Evaluating the integral at the limits of integration, we find the
displacement:
1
3(3)3−1
2(3)2−1
3(0)3−1
2(0)2=27
3−9
2
Step 7: Therefore, the displacement after 3 seconds is 27
3−9
2= 4.5 meters.
Question 16
Question
A car accelerates from rest along a straight road at 3.0 m/s2for 5.0 seconds. It
then maintains a constant velocity for 10 seconds before coming to a stop with
a constant deceleration of 2.0 m/s2. What is the total distance traveled by the
car during this entire motion?
14
Solution
Step 1: Find the distance traveled during acceleration. Given initial velocity
vi= 0 m/s, acceleration a= 3.0 m/s2, and time t= 5.0 s, we use the kinematic
equation:
d=vit+1
2at2
d= 0 + 1
2×3.0×(5.0)2
d= 0 + 22.5
d= 22.5 m
Step 2: Find the distance traveled at constant velocity. The velocity remains
constant for 10 seconds, so the distance traveled is:
d= velocity ×time = 3.0×10 = 30 m
Step 3: Find the distance traveled during deceleration. The car comes to
a stop with a constant deceleration of 2.0 m/s2. We need to find the distance
traveled during deceleration using the equation:
v2
f=v2
i+ 2ad
Since the final velocity vf= 0 m/s, the initial velocity vi= 3.0 m/s, the
deceleration a=−2.0 m/s2, we solve for d:
0 = (3.0)2+ 2 ×(−2.0) ×d
0=9−4d
4d= 9
d=9
4= 2.25 m
Step 4: Find the total distance traveled. The total distance traveled is the
sum of distances traveled during acceleration, constant velocity, and decelera-
tion:
dtotal = 22.5 + 30 + 2.25 = 54.75 m
Therefore, the total distance traveled by the car during this entire motion is
54.75 meters.
Question 17
Question
A car is initially at rest. It then accelerates along a straight road, such that
the velocity of the car at any time tis given by the function v(t) = 5t2−3t+ 2
where vis in m/s and tis in seconds. Determine the time tat which the car
comes to a stop.
15
Solution
Step 1: To find the time tat which the car comes to a stop, we need to determine
when the velocity v(t) is equal to zero.
Step 2: Set v(t) = 0:
5t2−3t+ 2 = 0
Step 3: Solve the quadratic equation by using the quadratic formula, t=
−b±√b2−4ac
2a, where a= 5, b=−3, and c= 2:
t=−(−3) ±p(−3)2−4·5·2
2·5
Step 4: Simplify the expression:
t=3±√9−40
10
Step 5: Since the discriminant is negative, there are no real solutions. This
means the car never comes to a stop along the road.
Question 18
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2for 8 seconds.
What is the final velocity of the car at the end of the 8-second interval?
Solution
Step 1: Identify the given variables.
Initial velocity vinitial = 0 m/s
Acceleration a= 2.5 m/s2
Time t= 8 s
Final velocity vfinal (unknown)
Step 2: Use the kinematic equation vfinal =vinitial +at to find the final
velocity.
Plugging in the given values, we get:
vfinal = 0 m/s + 2.5 m/s2×8 s
Step 3: Solve for the final velocity.
vfinal = 0 + 20 = 20 m/s
Answer: The final velocity of the car at the end of the 8-second interval is
20 m/s.
16
Question 19
Question
A car is traveling at a constant velocity of 25 m/s. At a certain instant, the
driver applies the brakes and comes to a stop after 50 m. If the deceleration of
the car is constant, what is the time it takes for the car to come to a stop?
Solution
Step 1: Identify the given variables and their values. The initial velocity of the
car, u= 25 m/s (since the car is traveling at a constant velocity). The final
velocity of the car, v= 0 (since the car comes to a stop). The displacement of
the car, s= 50 m. The acceleration of the car is given as the deceleration a,
which is opposite in direction to the motion of the car.
Step 2: Choose the appropriate kinematic equation to solve the problem.
The kinematic equation that relates initial velocity, final velocity, acceleration,
and displacement is:
v2=u2+ 2as
Step 3: Substitute the known values into the kinematic equation. Since the
final velocity is 0, the equation becomes:
02= (25 m/s)2+ 2a(50 m)
Step 4: Solve for the acceleration a.
0 = 625 + 100a
100a=−625
a=−6.25 m/s2
Step 5: Use the kinematic equation to find the time taken for the car to
stop. The kinematic equation relating initial velocity, acceleration, and time is:
v=u+at
Substitute the known values into the equation:
0 = 25 + (−6.25)t
6.25t= 25
t= 4 s
Therefore, it takes 4 seconds for the car to come to a stop.
17
Question 20
Question
A car accelerates uniformly from rest at 3.0 m/s
²
for 10 seconds, then maintains
a constant velocity for 20 seconds, and finally decelerates uniformly to a stop
in 5 seconds. Calculate the total distance traveled by the car during this entire
motion.
Solution
Step 1: Find the distance traveled during acceleration phase.
Given: Initial velocity, u= 0 m/s (car starts from rest) Acceleration, a= 3.0
m/s
²
Time, t= 10 s
Using the equation of motion: s=ut +1
2at2
Substitute the known values: s1= 0 ×10 + 1
2×3.0×(10)2s1= 0 + 0.5×
3.0×100 s1= 0 + 0.5×300 s1= 0 + 150 s1= 150 m
Therefore, the distance traveled during the acceleration phase is 150 m.
Step 2: Find the distance traveled during constant velocity phase.
Given: Velocity during constant velocity phase, v=u+at Constant velocity,
which means a= 0 Time, t= 20 s
Using the equation: s=vt
Substitute the known values: s2= 3.0×20 s2= 60 m
Therefore, the distance traveled during the constant velocity phase is 60 m.
Step 3: Find the distance traveled during deceleration phase.
Given: Final velocity, v= 0 m/s (car comes to a stop) Deceleration, which
is negative acceleration, a=−3.0 m/s
²
Time, t= 5 s
Using the equation of motion: s=ut +1
2at2
Substitute the known values: s3= 3.0×5 + 1
2×−3.0×(5)2s3= 15 −0.5×
3.0×25 s3= 15 −0.5×75 s3= 15 −37.5s3=−22.5 m
The negative sign indicates the direction of motion is opposite to the positive
direction.
Step 4: Calculate the total distance traveled by summing up the distances.
Total distance traveled, stotal =s1+s2+s3stotal = 150 + 60 −22.5stotal =
187.5 m
Therefore, the total distance traveled by the car during the entire motion is
187.5 m.
Question 21
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
It then maintains a constant velocity for 20 seconds before decelerating at a rate
of −3 m/s2until it comes to a stop. Find the total distance the car travels during
this entire process.
18
Solution
Step 1: Find the distance traveled during acceleration. The distance traveled
during acceleration can be found using the equation:
d=1
2·a·t2
where ais the acceleration and tis the time. Substitute a= 2 m/s2and t= 10 s:
d=1
2·2 m/s2·(10 s)2= 100 m
Step 2: Find the distance traveled during constant velocity. The distance
traveled during constant velocity can be found using the equation:
d=v·t
where vis the constant velocity and tis the time. The car maintains a constant
velocity for 20 seconds, so:
d=v·20
Since the car maintains a constant velocity during this time, the distance trav-
eled is simply the velocity multiplied by time. However, as the car starts from
rest and accelerates to this velocity, the distance traveled chronologically is the
sum of distances traveled during acceleration, constant velocity, and decelera-
tion.
Step 3: Find the distance traveled during deceleration. The distance traveled
during deceleration can also be found using the equation:
d=1
2·a·t2
Substitute a=−3 m/s2and the car comes to a stop, so the final velocity is 0.
0 = vf+a·t
t=−vf
a=0
−3= 0 s
So, the car comes to a stop immediately after moving at constant velocity for
20 seconds. The distance traveled during deceleration is:
d=1
2·(−3 m/s2)·(0 s)2= 0
Step 4: Calculate the total distance. The total distance traveled is the sum
of the distances traveled during acceleration, constant velocity, and deceleration:
Total distance = 100 m + v·20 + 0 = 100 m + v·20
Since the velocity is constant during the 20 seconds after acceleration, v=
2 m/s ·10 s = 20 m/s.
Total distance = 100 m + 20 m/s ·20 s = 100 m + 400 m = 500 m
Therefore, the total distance the car travels during this entire process is 500
meters.
19
Question 22
Question
A particle moves in a straight line with a velocity described by the equation
v(t) = 8 −4t+ 2t2, where vis the velocity in m/s and tis the time in seconds.
Determine the particle’s acceleration as a function of time.
Solution
Step 1: To find the particle’s acceleration as a function of time, we differentiate
the velocity function with respect to time. The acceleration a(t) is given by the
derivative of the velocity function v(t).
Step 1: a(t) = dv
dt
Step 2: Differentiate v(t) term by term:
Step 2: a(t) = d
dt(8 −4t+ 2t2)
Step 3: The derivative of a constant is 0, and the differentiation of −4tand
2t2with respect to tis −4 and 4t, respectively.
Step 3: a(t) = −4+4t
Step 4: Therefore, the particle’s acceleration as a function of time is a(t) =
−4 + 4t. This means that the acceleration is changing linearly with time, with
a constant acceleration of 4 m/s2.
Step 4: a(t) = −4+4t
Question 23
Question
An object is thrown straight up into the air from the ground with an initial
velocity of 20 m/s. How high above the ground does the object travel before
coming back down?
Solution
Step 1: Define the upward motion as positive.
Step 2: Identify the given quantities: Initial velocity, v0= 20 m/s
Acceleration due to gravity, a=−9.81 m/s2(negative because it acts downward)
Step 3: Use the kinematic equation for displacement to find the height the
object travels:
v2=v2
0+ 2a∆y
20
Substitute the given values:
0 = (20 m/s)2+ 2(−9.81 m/s2)∆y
Step 4: Solve for ∆y:
0 = 400 −19.62∆y
19.62∆y= 400
∆y=400
19.62
∆y≈20.38 m
Therefore, the object travels approximately 20.38 meters above the ground
before coming back down.
Question 24
Question
An object moves along a straight line according to the equation of motion x(t) =
5t3−3t2+2, where xis in meters and tis in seconds. Find the maximum velocity
of the object and the time at which it occurs.
Solution
Step 1: Find the velocity function by taking the derivative of the position
function with respect to time.
Step 1: v(t) = dx
dt =d
dt(5t3−3t2+ 2)
Step 2: Differentiate each term separately.
Step 2: v(t) = 15t2−6t
Step 3: To find the maximum velocity, we need to find the time when the
velocity is maximum. This occurs when the acceleration is zero. So, differentiate
the velocity function.
Step 3: a(t) = dv
dt =d
dt(15t2−6t)
Step 4: Differentiate each term separately.
Step 4: a(t) = 30t−6
Step 5: Set the acceleration function equal to zero and solve for t.
Step 5: 0 = 30t−6
21
30t= 6
t=6
30 = 0.2 s
Step 6: Find the maximum velocity by substituting t= 0.2 into the velocity
function.
Step 6: v(0.2) = 15(0.2)2−6(0.2)
v(0.2) = 15(0.04) −1.2
v(0.2) = 0.6−1.2 = −0.6 m/s
Therefore, the maximum velocity of the object is −0.6 m/s and it occurs at
t= 0.2 s.
Question 25
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 5.0
seconds. It then maintains a constant velocity for 10 seconds before coming
to a stop with a constant deceleration of 2.0 m/s2. What is the total distance
traveled by the car during this entire motion?
Solution
Step 1: Calculate the distance covered during acceleration. The distance covered
during acceleration can be calculated using the equation:
d=1
2at2
where ais the acceleration and tis the time taken. Plugging in the values, we
get:
d=1
2×3.0 m/s2×(5.0 s)2
d=1
2×3.0×25
d=75
2
d= 37.5 m
Step 2: Calculate the distance covered during constant velocity. The distance
covered during constant velocity can be calculated using the equation:
d=vt
where vis the constant velocity and tis the time taken. Given that the car
maintains a constant velocity, the distance covered is:
d=v×10
22
Since the car maintained a constant velocity, the distance covered is:
d= 0 ×10 = 0 m
Step 3: Calculate the distance covered during deceleration. The distance
covered during deceleration can be calculated using the equation:
d=1
2at2
where ais the deceleration and tis the time taken. Plugging in the values, we
get:
d=1
2×2.0 m/s2×(10 s)2
d=1
2×2.0× ×100
d= 100 m
Step 4: Calculate the total distance traveled. The total distance traveled
is the sum of the distances covered during acceleration, constant velocity, and
deceleration:
Total distance = 37.5 m + 0 m + 100 m
Total distance = 137.5 m
Therefore, the total distance traveled by the car during this entire motion is
137.5 meters.
Question 26
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2. Find the acceler-
ation of the car when the velocity is 24 m/s.
Solution
Step 1: Identify known values
The initial velocity of the car, u, is 0 m/s.
The final velocity of the car, v, is 24 m/s.
The acceleration of the car, a, is 2.0 m/s2.
Step 2: Choose the appropriate equation of motion
We can use the equation relating final velocity, initial velocity, acceleration, and
displacement:
v2=u2+ 2as
Step 3: Substitute known values into the equation
Substitute the known values into the equation:
23
(24 m/s)2= (0 m/s)2+ 2(2.0 m/s2)·s
Step 4: Solve for displacement (s)
576 m2/s2= 4.0 m/s2·s
s=576 m2/s2
4.0 m/s2= 144 m
Step 5: Calculate the acceleration at v= 24 m/s
Now that we have found the displacement (s), we can use the following equation
to find the acceleration (a) at v= 24 m/s:
v2=u2+ 2as
(24 m/s)2= (0 m/s)2+ 2a(144 m)
576 m2/s2= 288am
a=576 m2/s2
288 m = 2.0 m/s2
Therefore, the acceleration of the car when the velocity is 24 m/s is 2.0 m/s2.
Question 27
Question
A car starting from rest accelerates in a straight line at a constant rate of 2
m/s2. Calculate the time it takes for the car to reach a speed of 20 m/s.
Solution
Step 1: Identify the given values.
The initial velocity uof the car is 0 m/s, the acceleration ais 2 m/s2, and the
final velocity vis 20 m/s.
Step 2: Use the kinematic equation v=u+at
Substitute the known values into the equation:
v=u+at
20 = 0 + 2t
Step 3: Solve for time t.
20 = 2t
24
t=20
2
t= 10 seconds
Answer: It takes the car 10 seconds to reach a speed of 20 m/s.
Question 28
Question
A car traveling initially at a speed of 20 m/s accelerates uniformly at 2 m/s2for
10 seconds, then maintains a constant speed for the next 20 seconds. Calculate
the total distance the car travels during this time period.
Solution
Step 1: Calculate the distance covered during acceleration period. Given: Initial
velocity (u) = 20 m/s, acceleration (a) = 2 m/s2, time of acceleration (t) = 10
seconds.
Using the kinematic equation:
s=ut +1
2at2
We can substitute the given values to find the distance covered during ac-
celeration.
s= (20 m/s)(10 s) + 1
2(2 m/s2)(10 s)2
s= 200 m + 1
2(2)(100)
s= 200 m + 100
s= 300 m
Therefore, the distance covered during acceleration is 300 m.
Step 2: Calculate the distance covered during constant speed. During the
constant speed period, the car maintains a speed of 20 m/s for 20 seconds. The
distance covered during this time period is:
s= speed ×time = 20 m/s ×20 s = 400 m
Step 3: Calculate the total distance traveled. The total distance traveled is
the sum of the distances covered during acceleration and constant speed.
Total distance = 300 m + 400 m = 700 m
Therefore, the total distance the car travels during this time period is 700
meters.
25
Question 29
Question
A car starts from rest and accelerates along a straight road at 3.0 m/s2. At the
same instant, a truck passes the car traveling at a constant speed of 30 m/s.
How far down the road will the car overtake the truck?
Solution
Step 1: First, we find the time it takes for the car to overtake the truck. Let t
be the time it takes for the car to overtake the truck.
Step 2: We can set up equations for the motion of the car and the truck.
For the car, the position at time tis given by:
xcar =1
2at2
Step 3: For the truck, the position at time tis given by:
xtruck =vt
Step 4: When the car overtakes the truck, the positions are equal:
1
2at2=vt
Step 5: Substituting the given values for aand vinto the equation, we get:
1
2(3.0 m/s2)t2= (30 m/s)t
Step 6: Further simplifying the equation, we have:
1.5t2= 30t
Step 7: Dividing by t(assuming t= 0), we get:
1.5t= 30
t=30
1.5
t= 20 s
Step 8: Finally, we can find the distance the car travels to overtake the truck
by substituting tback into the equation for the car’s position:
xcar =1
2(3.0 m/s2)(20 s)2
xcar =1
2(3.0)(400) m
xcar = 600 m
Therefore, the car will overtake the truck after traveling 600 meters down
the road.
26
Question 30
Question
A car accelerates from rest at a constant rate of 3.0 m/s2. How far does the car
travel in the first 5.0 seconds?
Solution
Step 1: We will first find the final velocity of the car after 5.0 seconds using the
kinematic equation:
vf=vi+at
where vfis the final velocity, viis the initial velocity (in this case, 0 m/s), ais
the acceleration, and tis the time. Substituting the values:
vf= 0 + (3.0 m/s2)(5.0 s)
vf= 15.0 m/s
Step 2: Next, we will find the distance traveled by the car using the equation:
d=vit+1
2at2
where dis the distance traveled, viis the initial velocity, tis the time, and ais
the acceleration. Substituting the values:
d= 0 + 1
2(3.0 m/s2)(5.0 s)2
d=1
2(3.0 m/s2)(25 s2)
d=1
2(75 m)
d= 37.5 m
Therefore, the car travels a distance of 37.5 meters in the first 5.0 seconds.
Question 31
Question
A car initially at rest starts moving along a straight road. The car accelerates
uniformly over a distance of 200 meters and reaches a final velocity of 20 m/s.
Calculate the time it takes for the car to reach this final velocity.
27
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity (u) = 0 m/s (car is initially at rest)
Final velocity (v) = 20 m/s
Distance traveled (s) = 200 m
Acceleration (a) = ?
Time taken (t) = ?
Step 2: Use the kinematic equation relating displacement, initial velocity,
final velocity, acceleration, and time:
v=u+at
Step 3: Rearrange the equation to solve for acceleration:
a=v−u
t
Step 4: Since the car starts from rest, u= 0. Substitute the known values:
a=20 m/s −0 m/s
t
a=20 m/s
t
Step 5: Use the kinematic equation relating displacement, initial velocity,
acceleration, and time:
s=ut +1
2at2
Step 6: Substitute the known values:
200 m = 0 m/s ×t+1
2at2
200 = 1
2at2
400 = at2
Step 7: Substitute the expression for acceleration found in Step 4 into the
equation in Step 6:
400 = 20 m/s
t·t2
400 = 20t
t= 20 s
Therefore, it takes the car 20 seconds to reach a final velocity of 20 m/s.
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Question 32
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 10
seconds. What is the total distance traveled by the car during this time interval?
Solution
Step 1: Determine the final velocity of the car using the equation v=u+at,
where vis the final velocity, uis the initial velocity (0 m/s in this case), ais
the acceleration, and tis the time.
Final velocity = 0 + (3.0 m/s2)(10 s)
Final velocity = 30 m/s
Step 2: Calculate the total distance traveled by the car using the equation
for distance covered under constant acceleration, s=ut +1
2at2.
s= (0 m/s)(10 s) + 1
2(3.0 m/s2)(10 s)2
s= 0 + 0.5(3.0)(100)
s= 150 m
Therefore, the total distance traveled by the car during this time interval is
150 meters.
Question 33
Question
A car starts from rest and accelerates uniformly at 2.5 m/s2along a straight
track. How far has the car traveled when it reaches a speed of 25 m/s?
Solution
Step 1: Let’s denote the initial velocity of the car as vi= 0 m/s, the acceleration
as a= 2.5 m/s2, the final velocity as vf= 25 m/s, and the distance traveled as
d. We can use the following kinematic equation to relate these variables:
v2
f=v2
i+ 2ad
Step 2: Substituting the given values into the equation, we get:
(25 m/s)2= (0 m/s)2+ 2(2.5 m/s2)·d
Step 3: Simplifying the equation, we have:
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625 m/s2= 5 m/s2·d
Step 4: Solving for d, we find:
d=625 m/s2
5 m/s2= 125 m
So, the car has traveled 125 meters when it reaches a speed of 25 m/s.
Question 34
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the car decelerates at a constant rate of 4 m/s2until it comes
to a stop. Determine the total distance traveled by the car during this time
period.
Solution
Step 1: Let’s first determine the distance traveled during the acceleration phase.
The velocity of the car after 10 seconds can be found using the equation of
motion:
v=u+at,
where uis the initial velocity (0 m/s), ais the acceleration (2 m/s2), and tis
the time (10 seconds).
v= 0 + 2 ×10 = 20 m/s.
The distance traveled during the acceleration phase can be found using:
s=ut +1
2at2,
where sis the distance traveled.
s= 0 ×10 + 1
2×2×102=1
2×2×100 = 100 m.
Step 2: Next, let’s determine the distance traveled during the deceleration
phase. To find the time taken to stop during deceleration, we use:
v=u+at,
where uis the initial velocity (20 m/s), ais the deceleration (-4 m/s2), and vis
the final velocity (0 m/s).
0 = 20 −4t⇒t=20
4= 5 s.
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The distance traveled during deceleration can be found using:
s=ut +1
2at2,
where uis the initial velocity (20 m/s), tis the time (5 seconds), and ais the
deceleration (-4 m/s2).
s= 20 ×5 + 1
2× −4×52= 100 −50 = 50 m.
Step 3: Finally, the total distance traveled by the car is the sum of the
distances traveled during acceleration and deceleration.
Total distance = 100 m + 50 m = 150 m.
Therefore, the total distance traveled by the car during this time period is
150 meters.
Question 35
Question
A stone is thrown straight upward with an initial velocity of 20 m/s from the
edge of a cliff 100 m high. 1. How long will it take for the stone to hit the
ground at the foot of the cliff? 2. With what speed does it strike the ground?
Solution
Let’s first find the time it takes for the stone to hit the ground. We can use the
kinematic equation for vertical motion:
y=vit+1
2at2
where: y=−100 m (negative because down is taken as the negative direction),
vi= 20 m/s, a=−9.8 m/s2(acceleration due to gravity), and we want to find
t.
Step 1: Find the time it takes for the stone to hit the ground. Plugging in
the values, we have:
−100 = 20t−1
2·9.8·t2
Simplifying gives us:
4.9t2−20t−100 = 0
This is a quadratic equation where:
a= 4.9, b =−20, c =−100
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We can solve this equation using the quadratic formula:
t=−b±√b2−4ac
2a
Step 2: Using the quadratic formula to find t:
t=−(−20) ±p(−20)2−4·4.9·(−100)
2·4.9
t=20 ±√400 + 1960
9.8
t=20 ±√2360
9.8
Solving for tgives us two possible values: t= 10 s (time to go up) and
t= 4.08 s (time to fall back down). Since the time to hit the ground is the time
to go up plus the time to fall back down, the time taken for the stone to hit the
ground is t= 10 s + 4.08 s = 14.08 s.
Step 3: Now, let’s find the speed at which the stone strikes the ground. We
can use the equation for velocity in vertical motion:
v=vi+at
where: v= final velocity of the stone (we want to find this), vi= 20 m/s,
a=−9.8 m/s2(acceleration due to gravity), and t= 14.08 s.
Plugging in the values, we get:
v= 20 + (−9.8) ·14.08
v= 20 −138.08
v=−118.08 m/s
So, the stone strikes the ground with a speed of 118.08 m/s directed down-
ward.
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