PHYS 202 - GENERAL PHYSICS II -
Kinematics in One Dimension
Question Bank - Set 3
Liberty University
Question 1
Question
A car accelerates from rest along a straight road with a constant acceleration
of 3 m/s2. How long will it take for the car to reach a speed of 27 m/s?
Solution
Step 1: We can start by using the kinematic equation relating final velocity
(vf), initial velocity (vi), acceleration (a), and time (t):
vf=vi+at
Step 2: Since the car starts from rest, the initial velocity viis 0. Thus, the
equation simplifies to:
27 m/s = 0 + (3 m/s2)t
Step 3: Solving for t, we get:
t=27 m/s
3 m/s2= 9 s
Therefore, it will take 9 seconds for the car to reach a speed of 27 m/s.
Question 2
Question
A car starts from rest and accelerates uniformly at 2 m/s2along a straight road.
How long will it take for the car to reach a speed of 25 m/s?
Solution
Step 1: Identify the known variables and the unknown variable.
Let: - vf= 25 m/s (final velocity) - a= 2 m/s2(acceleration) - vi= 0 m/s
(initial velocity) - t(time taken)
Step 2: Choose the appropriate kinematic equation.
The kinematic equation that relates the final velocity, initial velocity, accelera-
tion, and time is:
vf=vi+at
Step 3: Plug in the known values into the chosen equation and solve for the
unknown.
Substitute the given values into the equation:
25 = 0 + 2t
Solve for t:
t=25
2= 12.5 s
Step 4: Check units and final answer.
The units for time are seconds (s). Hence, the car will take 12.5 seconds to
reach a speed of 25 m/s.
Question 3
Question
A car is traveling along a straight road with a velocity given by v(t) = 6t2−4t+5,
where vis in m/s and tis in seconds. Find the acceleration of the car at t= 2
seconds.
Solution
Step 1: To find the acceleration of the car, we need to differentiate the velocity
function with respect to time.
a(t) = dv
dt
Step 2: Given v(t)=6t2−4t+ 5, we can differentiate it with respect to t.
a(t) = d
dt(6t2−4t+ 5)
a(t) = 12t−4
Step 3: Now, we can find the acceleration at t= 2 seconds.
a(2) = 12(2) −4
a(2) = 24 −4
a(2) = 20 m/s2
Therefore, the acceleration of the car at t= 2 seconds is 20 m/s2.
2
Question 4
Question
A particle moves along a straight line. Its position (in meters) as a function of
time (in seconds) is given by the equation x(t) = 3t2−2t+ 4. Determine the
velocity and acceleration of the particle at time t= 2 s.
Solution
Step 1: To find the velocity of the particle at time t, we first need to find the
derivative of the position function x(t) with respect to time.
v(t) = dx
dt =d
dt(3t2−2t+ 4)
Step 2: Taking the derivative, we get:
v(t)=6t−2
Step 3: Now, to find the velocity at t= 2 s, we substitute t= 2 into the
velocity function:
v(2) = 6(2) −2 = 10 m/s
Step 4: Next, to find the acceleration of the particle, we take the derivative
of the velocity function v(t) with respect to time.
a(t) = dv
dt =d
dt(6t−2)
Step 5: Taking the derivative, we get:
a(t)=6
Step 6: Finally, to find the acceleration at t= 2 s, we substitute t= 2 into
the acceleration function:
a(2) = 6 m/s2
Therefore, the velocity of the particle at t= 2 s is 10 m/s and the acceleration
is 6 m/s2.
Question 5
Question
A car starts from rest and accelerates at a rate of 2 m/s2for 5 seconds. After
this time, the car maintains a constant velocity for another 10 seconds. Finally,
the car decelerates at a rate of 1 m/s2until it comes to a stop. Calculate the
total distance traveled by the car during this time.
3
Solution
Step 1: Calculate the distance traveled during acceleration phase. The distance
traveled during acceleration can be calculated using the equation:
d=1
2×a×t2
where ais the acceleration and tis the time taken.
Substitute a= 2 m/s2and t= 5 s into the equation:
d=1
2×2 m/s2×(5 s)2
d=1
2×2 m/s2×25 s2
d= 25 m
Therefore, the distance traveled during acceleration phase is 25 meters.
Step 2: Calculate the distance traveled during constant velocity phase. Since
the car maintains a constant velocity during this phase, the distance traveled is
simply:
d= velocity ×time
Since the car maintains a constant velocity, the distance traveled is given
by:
d= velocity ×time
Substitute the velocity during this phase:
d= 2 m/s ×10 s = 20 m
Therefore, the distance traveled during constant velocity phase is 20 meters.
Step 3: Calculate the distance traveled during deceleration phase. For de-
celeration, we can use the same kinematic equation as in acceleration phase:
d=1
2×a×t2
Substitute a=−1 m/s2and t= 15 s into the equation:
d=1
2×(−1 m/s2)×(15 s)2
d=1
2×(−1 m/s2)×225 s2
d=−112.5 m
Therefore, the distance traveled during deceleration phase is -112.5 meters.
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of distances during all phases:
Total distance = 25 m + 20 m −112.5 m = >32.5 m
Therefore, the total distance traveled by the car during this time is greater
than 32.5 meters.
4
Question 6
Question
A car accelerates from rest at a constant rate of 5 m/s2for a total of 10 seconds.
After this time, the car continues to move at a constant velocity. What is the
total distance traveled by the car in the first 20 seconds of motion?
Solution
Step 1: Find the distance traveled during the acceleration phase. The distance
traveled during acceleration can be found using the kinematic equation: d=
1
2at2, where dis the distance, ais the acceleration, and tis the time. Plugging
in the values, we have:
d=1
2×5 m/s2×(10 s)2
d=1
2×5 m/s2×100 s2
d= 250 m
Therefore, the car travels 250 meters during the acceleration phase.
Step 2: Find the distance traveled during the constant velocity phase. Since
the car is moving at a constant velocity, we can use the formula d=vt, where d
is the distance, vis the velocity, and tis the time. The velocity after 10 seconds
of acceleration would be:
v=a×t= 5 m/s2×10 s = 50 m/s
Using this constant velocity, the distance traveled in the next 10 seconds
would be:
d= 50 m/s ×10 s = 500 m
Step 3: Find the total distance traveled in the first 20 seconds. The total
distance traveled by the car in the first 20 seconds is the sum of the distances
traveled during acceleration and the constant velocity phase.
Total distance = 250 m + 500 m = 750 m
Therefore, the total distance traveled by the car in the first 20 seconds of
motion is 750 meters.
Question 7
Question
An object is initially at rest. It starts moving in a straight line with a constant
acceleration a, and after tseconds, its velocity is v. Find the displacement of
the object from its initial position after time t.
5
Solution
Step 1: We know that the object started from rest, so its initial velocity (v0) is
0 m/s.
Step 2: Using the kinematic equation v=v0+at, we can find the final
velocity of the object after time t:
v= 0 + at
v=at
Step 3: The average velocity of the object during time tis given by:
¯v=v0+v
2=0 + at
2=at
2
Step 4: The displacement of the object from its initial position can be found
using the definition of average velocity:
∆x= ¯v·t=at
2·t=1
2at2
Step 5: Therefore, the displacement of the object from its initial position
after time tis 1
2at2.
Question 8
Question
A car accelerates from rest at a constant rate of 2 m/s2for a distance of 50
meters. Determine the final velocity of the car.
Solution
Step 1: First, we need to find the final velocity of the car using the equation of
motion that relates the final velocity, initial velocity, acceleration, and distance:
v2
f=v2
i+ 2a∆x
where: vf= final velocity, vi= initial velocity (0 m/s since the car starts from
rest), a= acceleration (2 m/s2), ∆x= distance (50 m).
Step 2: Substitute the known values into the equation:
v2
f= (0 m/s)2+ 2(2 m/s2)(50 m)
Step 3: Simplify the equation:
v2
f= 0 + 2(2)(50) = 200 m2/s2
Step 4: Take the square root of both sides to solve for vf:
vf=√200 = 14.14 m/s
Therefore, the final velocity of the car is 14.14 m/s .
6
Question 9
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2for 5.0 seconds. It
then maintains a constant velocity for 10.0 seconds before applying the brakes,
coming to a stop in 4.0 seconds. Determine the distance the car travels during
the entire motion.
Solution
Step 1: Find the distance traveled during acceleration phase.
v=u+at
v= 3.0 m/s2×5.0 s
v= 15 m/s
The distance traveled during acceleration can be found using:
s=ut +1
2at2
s= 0 + 1
2×3.0 m/s2×(5.0 s)2
s=1
2×3.0 m/s2×25 s2
s= 37.5 m
Step 2: Find the distance traveled during the constant velocity phase. The
distance traveled during this phase is simply the product of the velocity and
time:
s= 15 m/s ×10 s
s= 150 m
Step 3: Find the distance traveled during deceleration phase. Since the car
comes to a stop, the distance traveled during deceleration can be found using
the equation:
v2=u2+ 2as
02= 152+ 2 ×(−3.0 m/s2)×s
s=152
2×3.0m
s=225
6m
s= 37.5 m
7
Step 4: Calculate the total distance traveled by summing up the distances
from each phase.
Total distance = 37.5 m + 150 m + 37.5 m
Total distance = 225 m
Therefore, the car travels a total distance of 225 meters during the entire
motion.
Question 10
Question
A car, initially traveling at a speed of 30 m/s, accelerates at a constant rate of
2 m/s2. How long will it take for the car to reach a speed of 50 m/s?
Solution
Step 1: Let’s set up the kinematic equation that relates final velocity (vf), initial
velocity (vi), acceleration (a), and time (t):
vf=vi+at
Step 2: Given that vi= 30 m/s, a= 2 m/s2, and vf= 50 m/s, we can
substitute these values into the equation:
50 = 30 + 2t
Step 3: Subtract 30 from both sides to isolate 2t:
50 −30 = 2t
20 = 2t
Step 4: Divide both sides by 2 to solve for t:
t=20
2
t= 10 s
Answer: It will take 10 seconds for the car to reach a speed of 50 m/s.
Question 11
Question
A car accelerates uniformly from rest along a straight road. After 5 seconds, it
reaches a speed of 20 m/s. Calculate the acceleration of the car.
8
Solution
Step 1: Identify known values and the acceleration formula. Let’s denote the
initial velocity of the car as vi= 0 m/s, the final velocity as vf= 20 m/s, the
time taken as t= 5 s, and the acceleration as a. The formula relating these
variables is:
vf=vi+a·t
Step 2: Plug in the known values and solve for acceleration. Substitute the
values into the formula:
20 = 0 + a·5
20 = 5a
a=20
5= 4 m/s2
Step 3: State the final answer. Therefore, the acceleration of the car is 4
m/s2.
Question 12
Question
A car moves along a straight road according to the equation x(t)=4t2−2t+ 1,
where xis in meters and tis in seconds. Determine the velocity of the car as a
function of time, and find the acceleration of the car when t= 3 seconds.
Solution
Step 1: To find the velocity of the car as a function of time, we need to differ-
entiate the position function x(t) with respect to time t:
v(t) = dx
dt =d(4t2−2t+ 1)
dt
Step 2: Differentiating each term separately, we get:
v(t)=8t−2
Step 3: Therefore, the velocity of the car as a function of time is v(t) = 8t−2
m/s.
Step 4: To find the acceleration of the car when t= 3 seconds, we need to
differentiate the velocity function v(t) with respect to time t:
a(t) = dv
dt =d(8t−2)
dt
Step 5: Differentiating each term, we get:
a(t)=8
Step 6: Therefore, the acceleration of the car when t= 3 seconds is a(3) = 8
m/s2.
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Question 13
Question
A car starts from rest and accelerates uniformly, reaching a speed of 20 m/s in
5 seconds. What is the acceleration of the car?
Solution
Step 1: Identify the known quantities: The initial velocity uis 0 m/s (the car
starts from rest), the final velocity vis 20 m/s, and the time tis 5 seconds.
Step 2: Recall the kinematic equation relating final velocity, initial velocity,
acceleration, and time:
v=u+at
where vis the final velocity, uis the initial velocity, ais the acceleration, and t
is the time.
Step 3: Substitute the known values into the equation:
20 = 0 + a×5
Step 4: Solve for the acceleration a:
20 = 5a
a=20
5
a= 4 m/s2
Step 5: The acceleration of the car is 4 m/s2.
Question 14
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 sec-
onds. After this time, the car maintains a constant velocity for 5 seconds before
decelerating at a rate of 3 m/s2until it comes to a stop. Determine the total
distance traveled during the entire motion.
Solution
Step 1: Find the distance covered during acceleration phase. The distance
covered during acceleration can be calculated using the equation:
d=1
2·a·t2
10
where ais the acceleration and tis the time. Substitute a= 2 m/s2and t= 10 s
into the formula:
d=1
2·2 m/s2·(10 s)2= 100 m
Step 2: Find the distance covered during constant velocity phase. The
distance covered during the constant velocity phase can be calculated using the
equation:
d=v·t
where vis the velocity and tis the time. Since the car maintains a constant
velocity during this phase, the distance covered is:
d=v·t= 20 m/s ·5 s = 100 m
Step 3: Find the distance covered during deceleration phase. The distance
covered during deceleration can be calculated using the equation:
d=1
2·a·t2
where ais the deceleration and tis the time. Substitute a= 3 m/s2and solve
for t, the time required to stop:
v=at
0 = 3 m/s2·t
t=0
3= 0 s
Since the time to stop is 0 seconds, the distance covered during deceleration is
also 0 meters.
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of the distances covered in each phase:
Total distance = 100 m + 100 m + 0 m = 200 m
Therefore, the total distance traveled during the entire motion is 200 meters.
Question 15
Question
A car starts from rest and accelerates uniformly at 2 m/s2for 10 seconds. After
this time, the car maintains a constant velocity for 5 seconds before decelerating
uniformly at 1 m/s2until it comes to a stop. Find the total distance traveled
by the car during this time.
11
Solution
Step 1: Find the distance covered during acceleration phase.
The final velocity at the end of acceleration can be found using the equation:
vfinal =vinitial +a·t
vfinal = 0 + 2 ·10 = 20 m/s
The distance covered during acceleration phase is given by:
Distance = 1
2·(vinitial +vfinal)·t
Distance = 1
2·(0 + 20) ·10 = 100 m
Step 2: Find the distance covered during uniform velocity phase.
During this phase, the car maintains a constant velocity of 20 m/s for 5
seconds. The distance covered during this phase is given by:
Distance = vaverage ·t
Distance = 20 ·5 = 100 m
Step 3: Find the distance covered during deceleration phase.
The final velocity at the end of deceleration can be found using the equation:
vfinal =vinitial +a·t
vfinal = 20 −1·t= 0
t=20
1= 20 s
The distance covered during deceleration phase is given by:
Distance = 1
2·(vinitial +vfinal)·t
Distance = 1
2·(20 + 0) ·20 = 200 m
Step 4: Find the total distance traveled by the car.
The total distance traveled is the sum of distances covered during each phase.
Total Distance = 100 + 100 + 200 = 400 m
Therefore, the total distance traveled by the car during this time is 400
meters.
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Question 16
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2. How long
will it take the car to reach a speed of 25 m/s?
Solution
Step 1: We will first identify the given quantities: Initial velocity, u= 0 m/s
Final velocity, v= 25 m/s
Acceleration, a= 2 m/s2
Time taken, t=?
Step 2: We can use the kinematic equation relating initial velocity, final
velocity, acceleration, and time:
v=u+at
Step 3: Substitute the given values into the equation:
25 = 0 + 2t
Step 4: Solve for t:
25 = 2t
t=25
2= 12.5 s
Step 5: Therefore, it will take the car 12.5 seconds to reach a speed of 25 m/s.
Question 17
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the driver applies the brakes, causing the car to decelerate at a
rate of 3 m/s2. How far does the car travel during the first 10 seconds? What
is the total distance traveled by the car until it comes to a stop?
Solution
Step 1: To find the distance traveled during the first 10 seconds while acceler-
ating, we can use the equation for displacement under constant acceleration:
s=vit+1
2at2
where s= displacement vi= initial velocity a= acceleration t= time
For the first 10 seconds: vi= 0 (starting from rest) a= 2 m/s2t= 10 s
13
Substitute these values into the equation:
s= 0 + 1
2×2×(10)2
s= 0 + 10 ×10
s= 100 m
Therefore, during the first 10 seconds, the car travels 100 meters.
Step 2: Next, we need to find the distance traveled during deceleration.
Since the car is coming to a stop due to deceleration, we can use the equation:
v2
f=v2
i+ 2a·s
where vf= final velocity s= displacement during deceleration
The final velocity at the end of acceleration is given by:
vf=vi+a·t
vf= 0 + 2 ·10
vf= 20 m/s
Using the equation for deceleration:
0 = (20)2+ 2 ·(−3) ·s
0 = 400 −6s
6s= 400
s=400
6
s= 66.6 m
Therefore, the car travels approximately 66.67 meters during deceleration.
Step 3: To find the total distance traveled by the car until it comes to a
stop, we simply add the distances traveled during acceleration and deceleration:
Total distance = 100 m + 66.6 m
Total distance ≈166.67 m
Thus, the total distance traveled by the car until it comes to a stop is ap-
proximately 166.67 meters.
Question 18
Question
A car is traveling along a straight road with a velocity given by v= (3.00t−
1.00t2) m/s, where tis in seconds. Determine the acceleration of the car when
t= 2.00 s.
14
Solution
Step 1: To find the acceleration of the car, we need to differentiate the velocity
function with respect to time. The acceleration is given by dv
dt .
Step 2: Differentiating v= 3.00t−1.00t2with respect to t, we get
a=dv
dt =d
dt(3.00t−1.00t2) = 3.00 −2.00t
Step 3: Substitute t= 2.00 s into the acceleration equation to find the
acceleration of the car at t= 2.00 s.
a= 3.00 −2.00(2.00) = 3.00 −4.00 = −1.00 m/s2
Therefore, the acceleration of the car when t= 2.00 s is −1.00 m/s2.
Question 19
Question
A car is traveling at a constant velocity of 30 m/s when the driver sees an ob-
stacle 150 m ahead. The driver applies the brakes, causing the car to decelerate
at a constant rate of 5 m/s2. What is the minimum reaction time required for
the driver to avoid hitting the obstacle?
Solution
Step 1: Let’s denote the time taken by the driver to react to the obstacle as
treact and the time taken by the car to come to a complete stop as tstop. We can
break down the problem into two parts: the time taken for the driver to react
and apply the brakes, and the time taken for the car to come to a stop after the
brakes are applied.
Step 2: The distance covered by the car during the reaction time can be
calculated using the formula for distance covered with constant velocity:
dreaction =vcar ·treact
Step 3: The distance remaining to the obstacle after the reaction time is
given by:
dremaining = 150 m −dreaction
Step 4: We can find the time taken for the car to come to a stop after the
brakes are applied using the equation of motion:
dremaining =vcartstop +1
2·a·t2
stop
Step 5: Substituting the known values into the above equation, we get:
150 m −30 m/s ·treact = 0 + 1
2·(−5 m/s2)·t2
stop
15
Step 6: Simplifying and rearranging the equation gives:
−2.5t2
stop = 150 m −30 m/s ·treact
Step 7: We also know that the total time taken is the sum of reaction time
and stop time:
ttotal =treact +tstop
Step 8: The minimum reaction time required is when the reaction time
is 0. This means the driver reacts instantaneously, so we can set treact to 0.
Substituting this into the total time equation gives:
ttotal = 0 + tstop =tstop
Step 9: Therefore, the minimum reaction time required for the driver to
avoid hitting the obstacle is the same as the time taken for the car to come to
a stop:
tstop =s150 m
2.5 m/s2≈7.75 s
So, the minimum reaction time required for the driver to avoid hitting the
obstacle is approximately 7.75 seconds.
Question 20
Question
A particle moves along the x-axis according to the equation x(t) = 12t2−2t3,
where xis in meters and tis in seconds. Find the velocity and acceleration of
the particle when t= 3 seconds.
Solution
Step 1: To find the velocity of the particle, we differentiate the position function
with respect to time.
Velocity, v(t) = dx
dt =d(12t2−2t3)
dt
v(t) = 24t−6t2
Step 2: To find the acceleration of the particle, we differentiate the velocity
function with respect to time.
Acceleration, a(t) = dv
dt =d(24t−6t2)
dt
a(t) = 24 −12t
16
Step 3: Substitute t= 3 seconds into the velocity function to find the velocity
at t= 3 seconds.
v(3) = 24(3) −6(3)2= 72 −54 = 18 m/s
Step 4: Substitute t= 3 seconds into the acceleration function to find the
acceleration at t= 3 seconds.
a(3) = 24 −12(3) = 24 −36 = −12 m/s2
Therefore, at t= 3 seconds, the velocity of the particle is 18 m/s and the
acceleration is -12 m/s2.
Question 21
Question
A car initially at rest accelerates along a straight road at a constant rate of
2 m/s2for 8 seconds. Determine the distance the car travels during this time
period.
Solution
Step 1: Identify the given variables and the unknown. Let vibe the initial
velocity of the car, abe the acceleration, tbe the time period, dbe the distance
traveled, and vfbe the final velocity of the car.
Step 2: Use the kinematic equation d=vit+1
2at2. Given that vi= 0,
a= 2 m/s2, and t= 8 s, we have:
d= 0 + 1
2×2×(8)2
Step 3: Calculate the distance traveled.
d= 0 + 1
2×2×64 = 64 m
Therefore, the car travels a distance of 64 meters during the 8-second time
period.
Question 22
Question
A particle starts from rest and moves along a straight line with a constant
acceleration of 3.5 m/s2. After 5 seconds, the particle’s velocity is 14 m/s. What
is the total distance covered by the particle during the first 5 seconds?
17
Solution
Step 1: Use the equation of motion relating velocity, initial velocity, acceleration,
and displacement:
v=u+at
where: - v= 14 m/s is the final velocity, - u= 0 m/s is the initial velocity, -
a= 3.5 m/s2is the acceleration, - t= 5 s is the time.
Step 2: Substitute the given values into the equation of motion to find the
displacement:
14 = 0 + 3.5×5
14 = 17.5
Displacement = 17.5 m
Step 3: The total distance covered by the particle during the first 5 seconds
is equal to the magnitude of the displacement. Therefore, the total distance
covered by the particle is 17.5 meters.
Question 23
Question
A car starts from rest and accelerates uniformly at 3 m/s2for 10 seconds. After
this time, the car maintains a constant speed for an additional 20 seconds.
Determine the total distance traveled by the car during this time interval.
Solution
Step 1: Calculate the distance covered during acceleration phase: The initial
velocity, vinitial, is 0 m/s. The acceleration, a, is 3 m/s2. The time, t, is 10
seconds.
The final velocity after 10 seconds of acceleration is given by:
vfinal =vinitial +a·t
vfinal = 0 + 3 ·10 = 30 m/s
The distance covered during acceleration is given by:
d1=1
2·(vinitial +vfinal)·t
d1=1
2·(0 + 30) ·10 = 150 m
Step 2: Calculate the distance during constant speed phase: The final veloc-
ity after acceleration phase is the initial velocity for the constant speed phase.
So, vinitial =vfinal = 30 m/s. The time for constant speed phase, t, is 20 seconds.
18
The distance covered during constant speed phase is given by:
d2=vinitial ·t
d2= 30 ·20 = 600 m
Step 3: Calculate the total distance traveled by the car: The total distance
traveled is the sum of the distances covered during acceleration and constant
speed phases.
Total distance = d1+d2= 150 + 600 = 750 m
Therefore, the total distance traveled by the car during this time interval is
750 meters.
Question 24
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2for 10 seconds. After
this time, the driver applies the brakes which produce a constant deceleration of
1.0 m/s2. Calculate the total distance the car travels before coming to a stop.
Solution
Step 1: Find the distance traveled during the acceleration phase. The final
velocity after 10 seconds of acceleration can be found using the kinematic equa-
tion:
vf=vi+at
vf= 0 + 2.0×10 = 20 m/s
The distance traveled during the acceleration phase can be calculated using
the kinematic equation:
d=vit+1
2at2
d= 0 ×10 + 1
2×2.0×(10)2= 100 m
Step 2: Find the distance traveled during the deceleration phase. The car
stops when its final velocity is 0. We can use the kinematic equation for uni-
formly decelerated motion to find the total distance traveled during deceleration:
v2
f=v2
i+ 2ad
0 = (20)2+ 2(−1.0)d
d=(20)2
2×1.0= 200 m
19
Step 3: Calculate the total distance traveled. The total distance traveled is
the sum of the distances traveled during acceleration and deceleration:
Total distance = 100 + 200 = 300 m
Therefore, the total distance the car travels before coming to a stop is 300
meters.
Question 25
Question
A car accelerates uniformly from rest to a speed of 24 m/s over a distance of
150 m. What is the magnitude of the car’s acceleration?
Solution
Step 1: Identify the given values. The final velocity of the car, vf, is 24 m/s;
the initial velocity of the car, vi, is 0 m/s; and the displacement of the car, d,
is 150 m.
Step 2: Use the kinematic equation v2
f=v2
i+ 2ad. Substitute the given
values into the equation to solve for the acceleration, a.
a=v2
f−v2
i
2d
a=(24 m/s)2−(0 m/s)2
2×150 m
a=576 m2/s2
300 m
a= 1.92 m/s2
Therefore, the magnitude of the car’s acceleration is 1.92 m/s2.
Question 26
Question
A car traveling along a straight road has a velocity of 12 m/s when the driver sees
a child in the road 50 m ahead. If the car’s maximum deceleration is −4 m/s2,
what is the maximum constant speed at which the car can travel without hitting
the child?
20
Solution
Step 1: Let’s assume that the car decelerates with its maximum deceleration,
a=−4 m/s2, until it stops just in time to avoid hitting the child. We can use
the kinematic equation v2
f=v2
i+ 2a∆x, where: - vfis the final velocity of the
car (which we want to find), - viis the initial velocity of the car (12 m/s), - ais
the acceleration of the car (−4 m/s2), - ∆xis the displacement of the car as it
decelerates (50 m).
Step 2: Substituting the given values into the kinematic equation:
v2
f= (12 m/s)2+ 2(−4 m/s2)(50 m)
v2
f= 144 m/s2−400 m/s2
v2
f=−256 m/s2
Step 3: Since the velocity cannot be negative in this context, the car can-
not stop in time with the maximum deceleration. Therefore, the car needs to
decelerate at a lower rate. Let’s assume the car decelerates with a constant
deceleration, a=−xm/s2(where x < 4), until it stops just in time to avoid
hitting the child.
Step 4: We apply the same kinematic equation: v2
f=v2
i+ 2a∆x, but with
the variable acceleration a=−x, to find the maximum speed that the car can
have.
Step 5: The final velocity, vf, should be zero when the car comes to a stop
just in time. Substituting the variables into the kinematic equation, we get:
0 = (12 m/s)2+ 2(−x)(50 m)
0 = 144 m/s2−100xm/s2
100x= 144
x=144
100 = 1.44 m/s2
Therefore, the maximum constant speed at which the car can travel without
hitting the child is 1.44 m/s.
Question 27
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2along a straight
road. Find the time it takes for the car to reach a speed of 25 m/s.
21
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, u= 0 m/s
Acceleration, a= 2.0 m/s2
Final velocity, v= 25 m/s
Unknown: Time taken, t
Step 2: Use the kinematic equation connecting v,u,a, and t.
The kinematic equation is:
v=u+at
Substitute the given values:
25 = 0 + 2.0t
Step 3: Solve for t.
25 = 2.0t
t=25
2.0
t= 12.5 s
Step 4: Check the answer.
We can also use another kinematic equation to check our answer:
v2=u2+ 2a(v−u)
Substitute the given values:
252= 0 + 2(2.0)(25 −0)
625 = 100
Since the equation holds true, the time calculated is correct.
Therefore, it takes the car 12.5 seconds to reach a speed of 25 m/s.
Question 28
Question
A particle moves along a straight line such that its position is given by x(t) =
3t3−14t2+ 22t−8, where xis in meters and tis in seconds. Find the particle’s
velocity at t= 2s.
22
Solution
Step 1: To find the velocity of the particle, we need to differentiate the position
function with respect to time.
Step 1: v(t) = dx
dt
Step 2: Differentiating x(t)=3t3−14t2+ 22t−8 with respect to tgives the
velocity function.
v(t) = dx
dt = 9t2−28t+ 22
Step 3: Substituting t= 2s into the velocity function gives the particle’s
velocity at t= 2s.
Step 3: v(2) = 9(2)2−28(2) + 22
v(2) = 36 −56 + 22
v(2) = 2 m/s
Therefore, the particle’s velocity at t= 2s is 2 m/s.
Question 29
Question
A car starts from rest and accelerates at 2.0 m/s2for 10 seconds. It then
maintains a constant velocity for 30 seconds before coming to a stop with a
uniform deceleration. If the total distance covered by the car during this entire
motion is 800 meters, what is the magnitude of the deceleration?
Solution
Step 1: Find the distance covered during acceleration phase
The distance covered during acceleration is given by the equation:
d=1
2at2
where ais the acceleration and tis the time. Substitute a= 2.0 m/s2and
t= 10 s:
dacceleration =1
2(2.0)(10)2= 100 m
Step 2: Find the distance covered during constant velocity phase
The car maintains a constant velocity for 30 seconds, so the distance covered
during this phase is:
dconstant velocity = velocity ×time
23
Since the car maintains constant velocity, the distance covered is:
dconstant velocity = velocity ×time = v×30
where vis the constant velocity.
Step 3: Find the distance covered during deceleration phase
The total distance covered is 800 meters. Therefore, the distance covered during
deceleration is:
ddeceleration = 800 −dacceleration −dconstant velocity
Step 4: Find the deceleration
The distance covered during deceleration phase is given by the equation:
d=1
2at2
where ais the deceleration and tis the time. The time taken to decelerate
from constant velocity to stop is the same as the acceleration time since the
acceleration and deceleration are uniform. Substitute t= 10 s:
ddeceleration =1
2a(10)2
Step 5: Solve for deceleration
Substitute the values for dacceleration,dconstant velocity, and ddeceleration into the
equation ddeceleration = 800 −dacceleration −dconstant velocity. Now solve for a, the
deceleration.
1
2a(10)2= 800 −100 −v×30
Solve for ausing this equation to find the magnitude of deceleration.
Question 30
Question
A car initially at rest starts moving along a straight road with a constant ac-
celeration of 2 m/s2. How far does the car travel in the first 5 seconds?
Solution
Step 1: We can use the equation of motion x=x0+v0t+1
2at2to find the
distance traveled by the car. Since the car starts from rest, v0= 0. Thus, the
equation simplifies to x=1
2at2.
Step 2: Substitute the values of aand tinto the equation:
x=1
2(2 m/s2)(5 s)2
24
Step 3: Calculate the distance traveled by the car:
x=1
2(2)(25) = 25 m
Therefore, the car travels 25 meters in the first 5 seconds.
Question 31
Question
A car initially at rest starts accelerating at a constant rate of 3 m/s2for 10
seconds. After this time, the car undergoes a constant deceleration of 2 m/s2. If
the car comes to a stop after a total time of 30 seconds, what is the maximum
speed reached by the car during this time?
Solution
Step 1: Find the distance traveled during acceleration.
The distance traveled during acceleration is given by the equation:
d1=1
2at2
where ais the acceleration and tis the time.
Substitute a= 3 m/s2and t= 10 s:
d1=1
2×3×(10)2= 150 m
Step 2: Find the distance traveled during deceleration.
The deceleration will stop the car, so the distance traveled during decel-
eration is the remaining distance. We find that by subtracting the distances
already traveled from the total distance.
Total distance, dtotal =d1+d2, where d1= 150 m (from step 1) and a=
2 m/s2(deceleration).
Let d2be the distance during deceleration:
d2=dtotal −d1= 150 −1
2×2×(30 −10)2
Solve for d2:
d2= 150 −1
2×2×202= 50 m
Step 3: Find the maximum speed.
The maximum speed is reached when the car transitions from acceleration
to deceleration. At this point, the speed remains constant. We can find this
maximum speed using the formula:
speed = initial speed + a×t
25
The initial speed is 0 m/s, acceleration is 3 m/s
²
for 10 seconds, so the speed
at the transition point is:
speed = 0 + 3 ×10 = 30 m/s
Therefore, the maximum speed reached by the car is 30 m/s .
Question 32
Question
A particle moves along a straight line such that its acceleration is given by
a(t)=6t−2 m/s2. If the particle is at rest at t= 1 s and its initial velocity is
3 m/s, determine the particle’s velocity at t= 3 s.
Solution
Step 1: To find the velocity function v(t), we will integrate the given acceleration
function a(t) with respect to time.
Za(t)dt =Z(6t−2) dt
Za(t)dt = 3t2−2t+C
where Cis the constant of integration.
Step 2: We know that the particle is at rest at t= 1 s, so at t= 1 s, v(1) = 0
m/s.
0 = 3(1)2−2(1) + C
0=3−2 + C
C=−1
Step 3: Therefore, the velocity function v(t) is given by:
v(t)=3t2−2t−1
Step 4: Now, to find the velocity at t= 3 s, we substitute t= 3 into the
velocity function v(t).
v(3) = 3(3)2−2(3) −1
v(3) = 27 −6−1
v(3) = 20 m/s
So, the particle’s velocity at t= 3 s is 20 m/s.
26
Question 33
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2for 10 seconds. How
far does the car travel during this time?
Solution
Step 1: We are given the initial velocity (u= 0 m/s), acceleration (a= 3.0 m/s2),
and time interval (t= 10 s). We need to find the distance traveled by the car
during this time. Since the car starts from rest, its initial velocity is 0 m/s.
Step 2: We can use the kinematic equation for displacement in terms of
initial velocity, acceleration, and time:
s=ut +1
2at2
Step 3: Substituting the given values into the equation, we have
s= 0 ×10 + 1
2×3.0×(10)2
Step 4: Simplifying the expression, we get
s= 0 + 1
2×3.0×100
Step 5: Therefore, the distance traveled by the car during this time is
s=1
2×3.0×100 = 150 m
Step 6: Hence, the car travels a distance of 150 meters during the 10-second
acceleration period.
Question 34
Question
A car initially at rest starts moving along a straight road with a constant accel-
eration of 2 m/s2. After 5 seconds, what is the car’s velocity and displacement
from its initial position?
Solution
Step 1: First, we find the car’s velocity after 5 seconds using the equation of
motion:
v=u+at
27
where vis the final velocity, uis the initial velocity, ais the acceleration, and t
is the time. Substitute u= 0m/s, a= 2m/s2, and t= 5s into the equation:
v= 0 + 2 ×5 = 10 m/s
Step 2: Next, we find the displacement of the car after 5 seconds using the
equation of motion:
s=ut +1
2at2
where sis the displacement. Substitute u= 0m/s, a= 2m/s2, and t= 5s into
the equation:
s= 0 ×5 + 1
2×2×52= 0 + 0 + 25 = 25 m
Therefore, after 5 seconds, the car’s velocity is 10 m/s and its displacement
from its initial position is 25 meters.
Question 35
Question
An object moves along a straight line with an initial velocity of 5 m/s and a
constant acceleration of -2 m/s2. What is the distance traveled by the object
when it first comes to rest?
Solution
Let’s denote the initial velocity as v0= 5 m/s, the acceleration as a=−2 m/s2,
the final velocity as vf= 0 m/s, and the distance traveled as d.
Step 1: Use the kinematic equation v2
f=v2
0+ 2ad to find the distance
traveled. Since vf= 0, we have:
0 = (5)2+ 2(−2)d
Step 2: Solve for d:
25 = −4d
d=−25
4
d=−6.25 m
Therefore, the object travels a distance of 6.25 meters before coming to rest.
28
Solution
Step 1: Identify the known variables and the unknown variable.
Let: - vf= 25 m/s (final velocity) - a= 2 m/s2(acceleration) - vi= 0 m/s
(initial velocity) - t(time taken)
Step 2: Choose the appropriate kinematic equation.
The kinematic equation that relates the final velocity, initial velocity, accelera-
tion, and time is:
vf=vi+at
Step 3: Plug in the known values into the chosen equation and solve for the
unknown.
Substitute the given values into the equation:
25 = 0 + 2t
Solve for t:
t=25
2= 12.5 s
Step 4: Check units and final answer.
The units for time are seconds (s). Hence, the car will take 12.5 seconds to
reach a speed of 25 m/s.
Question 3
Question
A car is traveling along a straight road with a velocity given by v(t) = 6t2−4t+5,
where vis in m/s and tis in seconds. Find the acceleration of the car at t= 2
seconds.
Solution
Step 1: To find the acceleration of the car, we need to differentiate the velocity
function with respect to time.
a(t) = dv
dt
Step 2: Given v(t)=6t2−4t+ 5, we can differentiate it with respect to t.
a(t) = d
dt(6t2−4t+ 5)
a(t) = 12t−4
Step 3: Now, we can find the acceleration at t= 2 seconds.
a(2) = 12(2) −4
a(2) = 24 −4
a(2) = 20 m/s2
Therefore, the acceleration of the car at t= 2 seconds is 20 m/s2.
2
Question 4
Question
A particle moves along a straight line. Its position (in meters) as a function of
time (in seconds) is given by the equation x(t) = 3t2−2t+ 4. Determine the
velocity and acceleration of the particle at time t= 2 s.
Solution
Step 1: To find the velocity of the particle at time t, we first need to find the
derivative of the position function x(t) with respect to time.
v(t) = dx
dt =d
dt(3t2−2t+ 4)
Step 2: Taking the derivative, we get:
v(t)=6t−2
Step 3: Now, to find the velocity at t= 2 s, we substitute t= 2 into the
velocity function:
v(2) = 6(2) −2 = 10 m/s
Step 4: Next, to find the acceleration of the particle, we take the derivative
of the velocity function v(t) with respect to time.
a(t) = dv
dt =d
dt(6t−2)
Step 5: Taking the derivative, we get:
a(t)=6
Step 6: Finally, to find the acceleration at t= 2 s, we substitute t= 2 into
the acceleration function:
a(2) = 6 m/s2
Therefore, the velocity of the particle at t= 2 s is 10 m/s and the acceleration
is 6 m/s2.
Question 5
Question
A car starts from rest and accelerates at a rate of 2 m/s2for 5 seconds. After
this time, the car maintains a constant velocity for another 10 seconds. Finally,
the car decelerates at a rate of 1 m/s2until it comes to a stop. Calculate the
total distance traveled by the car during this time.
3
Solution
Step 1: Calculate the distance traveled during acceleration phase. The distance
traveled during acceleration can be calculated using the equation:
d=1
2×a×t2
where ais the acceleration and tis the time taken.
Substitute a= 2 m/s2and t= 5 s into the equation:
d=1
2×2 m/s2×(5 s)2
d=1
2×2 m/s2×25 s2
d= 25 m
Therefore, the distance traveled during acceleration phase is 25 meters.
Step 2: Calculate the distance traveled during constant velocity phase. Since
the car maintains a constant velocity during this phase, the distance traveled is
simply:
d= velocity ×time
Since the car maintains a constant velocity, the distance traveled is given
by:
d= velocity ×time
Substitute the velocity during this phase:
d= 2 m/s ×10 s = 20 m
Therefore, the distance traveled during constant velocity phase is 20 meters.
Step 3: Calculate the distance traveled during deceleration phase. For de-
celeration, we can use the same kinematic equation as in acceleration phase:
d=1
2×a×t2
Substitute a=−1 m/s2and t= 15 s into the equation:
d=1
2×(−1 m/s2)×(15 s)2
d=1
2×(−1 m/s2)×225 s2
d=−112.5 m
Therefore, the distance traveled during deceleration phase is -112.5 meters.
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of distances during all phases:
Total distance = 25 m + 20 m −112.5 m = >32.5 m
Therefore, the total distance traveled by the car during this time is greater
than 32.5 meters.
4
Question 6
Question
A car accelerates from rest at a constant rate of 5 m/s2for a total of 10 seconds.
After this time, the car continues to move at a constant velocity. What is the
total distance traveled by the car in the first 20 seconds of motion?
Solution
Step 1: Find the distance traveled during the acceleration phase. The distance
traveled during acceleration can be found using the kinematic equation: d=
1
2at2, where dis the distance, ais the acceleration, and tis the time. Plugging
in the values, we have:
d=1
2×5 m/s2×(10 s)2
d=1
2×5 m/s2×100 s2
d= 250 m
Therefore, the car travels 250 meters during the acceleration phase.
Step 2: Find the distance traveled during the constant velocity phase. Since
the car is moving at a constant velocity, we can use the formula d=vt, where d
is the distance, vis the velocity, and tis the time. The velocity after 10 seconds
of acceleration would be:
v=a×t= 5 m/s2×10 s = 50 m/s
Using this constant velocity, the distance traveled in the next 10 seconds
would be:
d= 50 m/s ×10 s = 500 m
Step 3: Find the total distance traveled in the first 20 seconds. The total
distance traveled by the car in the first 20 seconds is the sum of the distances
traveled during acceleration and the constant velocity phase.
Total distance = 250 m + 500 m = 750 m
Therefore, the total distance traveled by the car in the first 20 seconds of
motion is 750 meters.
Question 7
Question
An object is initially at rest. It starts moving in a straight line with a constant
acceleration a, and after tseconds, its velocity is v. Find the displacement of
the object from its initial position after time t.
5
Solution
Step 1: We know that the object started from rest, so its initial velocity (v0) is
0 m/s.
Step 2: Using the kinematic equation v=v0+at, we can find the final
velocity of the object after time t:
v= 0 + at
v=at
Step 3: The average velocity of the object during time tis given by:
¯v=v0+v
2=0 + at
2=at
2
Step 4: The displacement of the object from its initial position can be found
using the definition of average velocity:
∆x= ¯v·t=at
2·t=1
2at2
Step 5: Therefore, the displacement of the object from its initial position
after time tis 1
2at2.
Question 8
Question
A car accelerates from rest at a constant rate of 2 m/s2for a distance of 50
meters. Determine the final velocity of the car.
Solution
Step 1: First, we need to find the final velocity of the car using the equation of
motion that relates the final velocity, initial velocity, acceleration, and distance:
v2
f=v2
i+ 2a∆x
where: vf= final velocity, vi= initial velocity (0 m/s since the car starts from
rest), a= acceleration (2 m/s2), ∆x= distance (50 m).
Step 2: Substitute the known values into the equation:
v2
f= (0 m/s)2+ 2(2 m/s2)(50 m)
Step 3: Simplify the equation:
v2
f= 0 + 2(2)(50) = 200 m2/s2
Step 4: Take the square root of both sides to solve for vf:
vf=√200 = 14.14 m/s
Therefore, the final velocity of the car is 14.14 m/s .
6
Question 9
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2for 5.0 seconds. It
then maintains a constant velocity for 10.0 seconds before applying the brakes,
coming to a stop in 4.0 seconds. Determine the distance the car travels during
the entire motion.
Solution
Step 1: Find the distance traveled during acceleration phase.
v=u+at
v= 3.0 m/s2×5.0 s
v= 15 m/s
The distance traveled during acceleration can be found using:
s=ut +1
2at2
s= 0 + 1
2×3.0 m/s2×(5.0 s)2
s=1
2×3.0 m/s2×25 s2
s= 37.5 m
Step 2: Find the distance traveled during the constant velocity phase. The
distance traveled during this phase is simply the product of the velocity and
time:
s= 15 m/s ×10 s
s= 150 m
Step 3: Find the distance traveled during deceleration phase. Since the car
comes to a stop, the distance traveled during deceleration can be found using
the equation:
v2=u2+ 2as
02= 152+ 2 ×(−3.0 m/s2)×s
s=152
2×3.0m
s=225
6m
s= 37.5 m
7
Step 4: Calculate the total distance traveled by summing up the distances
from each phase.
Total distance = 37.5 m + 150 m + 37.5 m
Total distance = 225 m
Therefore, the car travels a total distance of 225 meters during the entire
motion.
Question 10
Question
A car, initially traveling at a speed of 30 m/s, accelerates at a constant rate of
2 m/s2. How long will it take for the car to reach a speed of 50 m/s?
Solution
Step 1: Let’s set up the kinematic equation that relates final velocity (vf), initial
velocity (vi), acceleration (a), and time (t):
vf=vi+at
Step 2: Given that vi= 30 m/s, a= 2 m/s2, and vf= 50 m/s, we can
substitute these values into the equation:
50 = 30 + 2t
Step 3: Subtract 30 from both sides to isolate 2t:
50 −30 = 2t
20 = 2t
Step 4: Divide both sides by 2 to solve for t:
t=20
2
t= 10 s
Answer: It will take 10 seconds for the car to reach a speed of 50 m/s.
Question 11
Question
A car accelerates uniformly from rest along a straight road. After 5 seconds, it
reaches a speed of 20 m/s. Calculate the acceleration of the car.
8
Solution
Step 1: Identify known values and the acceleration formula. Let’s denote the
initial velocity of the car as vi= 0 m/s, the final velocity as vf= 20 m/s, the
time taken as t= 5 s, and the acceleration as a. The formula relating these
variables is:
vf=vi+a·t
Step 2: Plug in the known values and solve for acceleration. Substitute the
values into the formula:
20 = 0 + a·5
20 = 5a
a=20
5= 4 m/s2
Step 3: State the final answer. Therefore, the acceleration of the car is 4
m/s2.
Question 12
Question
A car moves along a straight road according to the equation x(t)=4t2−2t+ 1,
where xis in meters and tis in seconds. Determine the velocity of the car as a
function of time, and find the acceleration of the car when t= 3 seconds.
Solution
Step 1: To find the velocity of the car as a function of time, we need to differ-
entiate the position function x(t) with respect to time t:
v(t) = dx
dt =d(4t2−2t+ 1)
dt
Step 2: Differentiating each term separately, we get:
v(t)=8t−2
Step 3: Therefore, the velocity of the car as a function of time is v(t) = 8t−2
m/s.
Step 4: To find the acceleration of the car when t= 3 seconds, we need to
differentiate the velocity function v(t) with respect to time t:
a(t) = dv
dt =d(8t−2)
dt
Step 5: Differentiating each term, we get:
a(t)=8
Step 6: Therefore, the acceleration of the car when t= 3 seconds is a(3) = 8
m/s2.
9
Question 13
Question
A car starts from rest and accelerates uniformly, reaching a speed of 20 m/s in
5 seconds. What is the acceleration of the car?
Solution
Step 1: Identify the known quantities: The initial velocity uis 0 m/s (the car
starts from rest), the final velocity vis 20 m/s, and the time tis 5 seconds.
Step 2: Recall the kinematic equation relating final velocity, initial velocity,
acceleration, and time:
v=u+at
where vis the final velocity, uis the initial velocity, ais the acceleration, and t
is the time.
Step 3: Substitute the known values into the equation:
20 = 0 + a×5
Step 4: Solve for the acceleration a:
20 = 5a
a=20
5
a= 4 m/s2
Step 5: The acceleration of the car is 4 m/s2.
Question 14
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 sec-
onds. After this time, the car maintains a constant velocity for 5 seconds before
decelerating at a rate of 3 m/s2until it comes to a stop. Determine the total
distance traveled during the entire motion.
Solution
Step 1: Find the distance covered during acceleration phase. The distance
covered during acceleration can be calculated using the equation:
d=1
2·a·t2
10
where ais the acceleration and tis the time. Substitute a= 2 m/s2and t= 10 s
into the formula:
d=1
2·2 m/s2·(10 s)2= 100 m
Step 2: Find the distance covered during constant velocity phase. The
distance covered during the constant velocity phase can be calculated using the
equation:
d=v·t
where vis the velocity and tis the time. Since the car maintains a constant
velocity during this phase, the distance covered is:
d=v·t= 20 m/s ·5 s = 100 m
Step 3: Find the distance covered during deceleration phase. The distance
covered during deceleration can be calculated using the equation:
d=1
2·a·t2
where ais the deceleration and tis the time. Substitute a= 3 m/s2and solve
for t, the time required to stop:
v=at
0 = 3 m/s2·t
t=0
3= 0 s
Since the time to stop is 0 seconds, the distance covered during deceleration is
also 0 meters.
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of the distances covered in each phase:
Total distance = 100 m + 100 m + 0 m = 200 m
Therefore, the total distance traveled during the entire motion is 200 meters.
Question 15
Question
A car starts from rest and accelerates uniformly at 2 m/s2for 10 seconds. After
this time, the car maintains a constant velocity for 5 seconds before decelerating
uniformly at 1 m/s2until it comes to a stop. Find the total distance traveled
by the car during this time.
11
Solution
Step 1: Find the distance covered during acceleration phase.
The final velocity at the end of acceleration can be found using the equation:
vfinal =vinitial +a·t
vfinal = 0 + 2 ·10 = 20 m/s
The distance covered during acceleration phase is given by:
Distance = 1
2·(vinitial +vfinal)·t
Distance = 1
2·(0 + 20) ·10 = 100 m
Step 2: Find the distance covered during uniform velocity phase.
During this phase, the car maintains a constant velocity of 20 m/s for 5
seconds. The distance covered during this phase is given by:
Distance = vaverage ·t
Distance = 20 ·5 = 100 m
Step 3: Find the distance covered during deceleration phase.
The final velocity at the end of deceleration can be found using the equation:
vfinal =vinitial +a·t
vfinal = 20 −1·t= 0
t=20
1= 20 s
The distance covered during deceleration phase is given by:
Distance = 1
2·(vinitial +vfinal)·t
Distance = 1
2·(20 + 0) ·20 = 200 m
Step 4: Find the total distance traveled by the car.
The total distance traveled is the sum of distances covered during each phase.
Total Distance = 100 + 100 + 200 = 400 m
Therefore, the total distance traveled by the car during this time is 400
meters.
12
Question 16
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2. How long
will it take the car to reach a speed of 25 m/s?
Solution
Step 1: We will first identify the given quantities: Initial velocity, u= 0 m/s
Final velocity, v= 25 m/s
Acceleration, a= 2 m/s2
Time taken, t=?
Step 2: We can use the kinematic equation relating initial velocity, final
velocity, acceleration, and time:
v=u+at
Step 3: Substitute the given values into the equation:
25 = 0 + 2t
Step 4: Solve for t:
25 = 2t
t=25
2= 12.5 s
Step 5: Therefore, it will take the car 12.5 seconds to reach a speed of 25 m/s.
Question 17
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2for 10 seconds.
After this time, the driver applies the brakes, causing the car to decelerate at a
rate of 3 m/s2. How far does the car travel during the first 10 seconds? What
is the total distance traveled by the car until it comes to a stop?
Solution
Step 1: To find the distance traveled during the first 10 seconds while acceler-
ating, we can use the equation for displacement under constant acceleration:
s=vit+1
2at2
where s= displacement vi= initial velocity a= acceleration t= time
For the first 10 seconds: vi= 0 (starting from rest) a= 2 m/s2t= 10 s
13
Substitute these values into the equation:
s= 0 + 1
2×2×(10)2
s= 0 + 10 ×10
s= 100 m
Therefore, during the first 10 seconds, the car travels 100 meters.
Step 2: Next, we need to find the distance traveled during deceleration.
Since the car is coming to a stop due to deceleration, we can use the equation:
v2
f=v2
i+ 2a·s
where vf= final velocity s= displacement during deceleration
The final velocity at the end of acceleration is given by:
vf=vi+a·t
vf= 0 + 2 ·10
vf= 20 m/s
Using the equation for deceleration:
0 = (20)2+ 2 ·(−3) ·s
0 = 400 −6s
6s= 400
s=400
6
s= 66.6 m
Therefore, the car travels approximately 66.67 meters during deceleration.
Step 3: To find the total distance traveled by the car until it comes to a
stop, we simply add the distances traveled during acceleration and deceleration:
Total distance = 100 m + 66.6 m
Total distance ≈166.67 m
Thus, the total distance traveled by the car until it comes to a stop is ap-
proximately 166.67 meters.
Question 18
Question
A car is traveling along a straight road with a velocity given by v= (3.00t−
1.00t2) m/s, where tis in seconds. Determine the acceleration of the car when
t= 2.00 s.
14
Solution
Step 1: To find the acceleration of the car, we need to differentiate the velocity
function with respect to time. The acceleration is given by dv
dt .
Step 2: Differentiating v= 3.00t−1.00t2with respect to t, we get
a=dv
dt =d
dt(3.00t−1.00t2) = 3.00 −2.00t
Step 3: Substitute t= 2.00 s into the acceleration equation to find the
acceleration of the car at t= 2.00 s.
a= 3.00 −2.00(2.00) = 3.00 −4.00 = −1.00 m/s2
Therefore, the acceleration of the car when t= 2.00 s is −1.00 m/s2.
Question 19
Question
A car is traveling at a constant velocity of 30 m/s when the driver sees an ob-
stacle 150 m ahead. The driver applies the brakes, causing the car to decelerate
at a constant rate of 5 m/s2. What is the minimum reaction time required for
the driver to avoid hitting the obstacle?
Solution
Step 1: Let’s denote the time taken by the driver to react to the obstacle as
treact and the time taken by the car to come to a complete stop as tstop. We can
break down the problem into two parts: the time taken for the driver to react
and apply the brakes, and the time taken for the car to come to a stop after the
brakes are applied.
Step 2: The distance covered by the car during the reaction time can be
calculated using the formula for distance covered with constant velocity:
dreaction =vcar ·treact
Step 3: The distance remaining to the obstacle after the reaction time is
given by:
dremaining = 150 m −dreaction
Step 4: We can find the time taken for the car to come to a stop after the
brakes are applied using the equation of motion:
dremaining =vcartstop +1
2·a·t2
stop
Step 5: Substituting the known values into the above equation, we get:
150 m −30 m/s ·treact = 0 + 1
2·(−5 m/s2)·t2
stop
15
Step 6: Simplifying and rearranging the equation gives:
−2.5t2
stop = 150 m −30 m/s ·treact
Step 7: We also know that the total time taken is the sum of reaction time
and stop time:
ttotal =treact +tstop
Step 8: The minimum reaction time required is when the reaction time
is 0. This means the driver reacts instantaneously, so we can set treact to 0.
Substituting this into the total time equation gives:
ttotal = 0 + tstop =tstop
Step 9: Therefore, the minimum reaction time required for the driver to
avoid hitting the obstacle is the same as the time taken for the car to come to
a stop:
tstop =s150 m
2.5 m/s2≈7.75 s
So, the minimum reaction time required for the driver to avoid hitting the
obstacle is approximately 7.75 seconds.
Question 20
Question
A particle moves along the x-axis according to the equation x(t) = 12t2−2t3,
where xis in meters and tis in seconds. Find the velocity and acceleration of
the particle when t= 3 seconds.
Solution
Step 1: To find the velocity of the particle, we differentiate the position function
with respect to time.
Velocity, v(t) = dx
dt =d(12t2−2t3)
dt
v(t) = 24t−6t2
Step 2: To find the acceleration of the particle, we differentiate the velocity
function with respect to time.
Acceleration, a(t) = dv
dt =d(24t−6t2)
dt
a(t) = 24 −12t
16
Step 3: Substitute t= 3 seconds into the velocity function to find the velocity
at t= 3 seconds.
v(3) = 24(3) −6(3)2= 72 −54 = 18 m/s
Step 4: Substitute t= 3 seconds into the acceleration function to find the
acceleration at t= 3 seconds.
a(3) = 24 −12(3) = 24 −36 = −12 m/s2
Therefore, at t= 3 seconds, the velocity of the particle is 18 m/s and the
acceleration is -12 m/s2.
Question 21
Question
A car initially at rest accelerates along a straight road at a constant rate of
2 m/s2for 8 seconds. Determine the distance the car travels during this time
period.
Solution
Step 1: Identify the given variables and the unknown. Let vibe the initial
velocity of the car, abe the acceleration, tbe the time period, dbe the distance
traveled, and vfbe the final velocity of the car.
Step 2: Use the kinematic equation d=vit+1
2at2. Given that vi= 0,
a= 2 m/s2, and t= 8 s, we have:
d= 0 + 1
2×2×(8)2
Step 3: Calculate the distance traveled.
d= 0 + 1
2×2×64 = 64 m
Therefore, the car travels a distance of 64 meters during the 8-second time
period.
Question 22
Question
A particle starts from rest and moves along a straight line with a constant
acceleration of 3.5 m/s2. After 5 seconds, the particle’s velocity is 14 m/s. What
is the total distance covered by the particle during the first 5 seconds?
17
Solution
Step 1: Use the equation of motion relating velocity, initial velocity, acceleration,
and displacement:
v=u+at
where: - v= 14 m/s is the final velocity, - u= 0 m/s is the initial velocity, -
a= 3.5 m/s2is the acceleration, - t= 5 s is the time.
Step 2: Substitute the given values into the equation of motion to find the
displacement:
14 = 0 + 3.5×5
14 = 17.5
Displacement = 17.5 m
Step 3: The total distance covered by the particle during the first 5 seconds
is equal to the magnitude of the displacement. Therefore, the total distance
covered by the particle is 17.5 meters.
Question 23
Question
A car starts from rest and accelerates uniformly at 3 m/s2for 10 seconds. After
this time, the car maintains a constant speed for an additional 20 seconds.
Determine the total distance traveled by the car during this time interval.
Solution
Step 1: Calculate the distance covered during acceleration phase: The initial
velocity, vinitial, is 0 m/s. The acceleration, a, is 3 m/s2. The time, t, is 10
seconds.
The final velocity after 10 seconds of acceleration is given by:
vfinal =vinitial +a·t
vfinal = 0 + 3 ·10 = 30 m/s
The distance covered during acceleration is given by:
d1=1
2·(vinitial +vfinal)·t
d1=1
2·(0 + 30) ·10 = 150 m
Step 2: Calculate the distance during constant speed phase: The final veloc-
ity after acceleration phase is the initial velocity for the constant speed phase.
So, vinitial =vfinal = 30 m/s. The time for constant speed phase, t, is 20 seconds.
18
The distance covered during constant speed phase is given by:
d2=vinitial ·t
d2= 30 ·20 = 600 m
Step 3: Calculate the total distance traveled by the car: The total distance
traveled is the sum of the distances covered during acceleration and constant
speed phases.
Total distance = d1+d2= 150 + 600 = 750 m
Therefore, the total distance traveled by the car during this time interval is
750 meters.
Question 24
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2for 10 seconds. After
this time, the driver applies the brakes which produce a constant deceleration of
1.0 m/s2. Calculate the total distance the car travels before coming to a stop.
Solution
Step 1: Find the distance traveled during the acceleration phase. The final
velocity after 10 seconds of acceleration can be found using the kinematic equa-
tion:
vf=vi+at
vf= 0 + 2.0×10 = 20 m/s
The distance traveled during the acceleration phase can be calculated using
the kinematic equation:
d=vit+1
2at2
d= 0 ×10 + 1
2×2.0×(10)2= 100 m
Step 2: Find the distance traveled during the deceleration phase. The car
stops when its final velocity is 0. We can use the kinematic equation for uni-
formly decelerated motion to find the total distance traveled during deceleration:
v2
f=v2
i+ 2ad
0 = (20)2+ 2(−1.0)d
d=(20)2
2×1.0= 200 m
19
Step 3: Calculate the total distance traveled. The total distance traveled is
the sum of the distances traveled during acceleration and deceleration:
Total distance = 100 + 200 = 300 m
Therefore, the total distance the car travels before coming to a stop is 300
meters.
Question 25
Question
A car accelerates uniformly from rest to a speed of 24 m/s over a distance of
150 m. What is the magnitude of the car’s acceleration?
Solution
Step 1: Identify the given values. The final velocity of the car, vf, is 24 m/s;
the initial velocity of the car, vi, is 0 m/s; and the displacement of the car, d,
is 150 m.
Step 2: Use the kinematic equation v2
f=v2
i+ 2ad. Substitute the given
values into the equation to solve for the acceleration, a.
a=v2
f−v2
i
2d
a=(24 m/s)2−(0 m/s)2
2×150 m
a=576 m2/s2
300 m
a= 1.92 m/s2
Therefore, the magnitude of the car’s acceleration is 1.92 m/s2.
Question 26
Question
A car traveling along a straight road has a velocity of 12 m/s when the driver sees
a child in the road 50 m ahead. If the car’s maximum deceleration is −4 m/s2,
what is the maximum constant speed at which the car can travel without hitting
the child?
20
Solution
Step 1: Let’s assume that the car decelerates with its maximum deceleration,
a=−4 m/s2, until it stops just in time to avoid hitting the child. We can use
the kinematic equation v2
f=v2
i+ 2a∆x, where: - vfis the final velocity of the
car (which we want to find), - viis the initial velocity of the car (12 m/s), - ais
the acceleration of the car (−4 m/s2), - ∆xis the displacement of the car as it
decelerates (50 m).
Step 2: Substituting the given values into the kinematic equation:
v2
f= (12 m/s)2+ 2(−4 m/s2)(50 m)
v2
f= 144 m/s2−400 m/s2
v2
f=−256 m/s2
Step 3: Since the velocity cannot be negative in this context, the car can-
not stop in time with the maximum deceleration. Therefore, the car needs to
decelerate at a lower rate. Let’s assume the car decelerates with a constant
deceleration, a=−xm/s2(where x < 4), until it stops just in time to avoid
hitting the child.
Step 4: We apply the same kinematic equation: v2
f=v2
i+ 2a∆x, but with
the variable acceleration a=−x, to find the maximum speed that the car can
have.
Step 5: The final velocity, vf, should be zero when the car comes to a stop
just in time. Substituting the variables into the kinematic equation, we get:
0 = (12 m/s)2+ 2(−x)(50 m)
0 = 144 m/s2−100xm/s2
100x= 144
x=144
100 = 1.44 m/s2
Therefore, the maximum constant speed at which the car can travel without
hitting the child is 1.44 m/s.
Question 27
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2along a straight
road. Find the time it takes for the car to reach a speed of 25 m/s.
21
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, u= 0 m/s
Acceleration, a= 2.0 m/s2
Final velocity, v= 25 m/s
Unknown: Time taken, t
Step 2: Use the kinematic equation connecting v,u,a, and t.
The kinematic equation is:
v=u+at
Substitute the given values:
25 = 0 + 2.0t
Step 3: Solve for t.
25 = 2.0t
t=25
2.0
t= 12.5 s
Step 4: Check the answer.
We can also use another kinematic equation to check our answer:
v2=u2+ 2a(v−u)
Substitute the given values:
252= 0 + 2(2.0)(25 −0)
625 = 100
Since the equation holds true, the time calculated is correct.
Therefore, it takes the car 12.5 seconds to reach a speed of 25 m/s.
Question 28
Question
A particle moves along a straight line such that its position is given by x(t) =
3t3−14t2+ 22t−8, where xis in meters and tis in seconds. Find the particle’s
velocity at t= 2s.
22
Solution
Step 1: To find the velocity of the particle, we need to differentiate the position
function with respect to time.
Step 1: v(t) = dx
dt
Step 2: Differentiating x(t)=3t3−14t2+ 22t−8 with respect to tgives the
velocity function.
v(t) = dx
dt = 9t2−28t+ 22
Step 3: Substituting t= 2s into the velocity function gives the particle’s
velocity at t= 2s.
Step 3: v(2) = 9(2)2−28(2) + 22
v(2) = 36 −56 + 22
v(2) = 2 m/s
Therefore, the particle’s velocity at t= 2s is 2 m/s.
Question 29
Question
A car starts from rest and accelerates at 2.0 m/s2for 10 seconds. It then
maintains a constant velocity for 30 seconds before coming to a stop with a
uniform deceleration. If the total distance covered by the car during this entire
motion is 800 meters, what is the magnitude of the deceleration?
Solution
Step 1: Find the distance covered during acceleration phase
The distance covered during acceleration is given by the equation:
d=1
2at2
where ais the acceleration and tis the time. Substitute a= 2.0 m/s2and
t= 10 s:
dacceleration =1
2(2.0)(10)2= 100 m
Step 2: Find the distance covered during constant velocity phase
The car maintains a constant velocity for 30 seconds, so the distance covered
during this phase is:
dconstant velocity = velocity ×time
23
Since the car maintains constant velocity, the distance covered is:
dconstant velocity = velocity ×time = v×30
where vis the constant velocity.
Step 3: Find the distance covered during deceleration phase
The total distance covered is 800 meters. Therefore, the distance covered during
deceleration is:
ddeceleration = 800 −dacceleration −dconstant velocity
Step 4: Find the deceleration
The distance covered during deceleration phase is given by the equation:
d=1
2at2
where ais the deceleration and tis the time. The time taken to decelerate
from constant velocity to stop is the same as the acceleration time since the
acceleration and deceleration are uniform. Substitute t= 10 s:
ddeceleration =1
2a(10)2
Step 5: Solve for deceleration
Substitute the values for dacceleration,dconstant velocity, and ddeceleration into the
equation ddeceleration = 800 −dacceleration −dconstant velocity. Now solve for a, the
deceleration.
1
2a(10)2= 800 −100 −v×30
Solve for ausing this equation to find the magnitude of deceleration.
Question 30
Question
A car initially at rest starts moving along a straight road with a constant ac-
celeration of 2 m/s2. How far does the car travel in the first 5 seconds?
Solution
Step 1: We can use the equation of motion x=x0+v0t+1
2at2to find the
distance traveled by the car. Since the car starts from rest, v0= 0. Thus, the
equation simplifies to x=1
2at2.
Step 2: Substitute the values of aand tinto the equation:
x=1
2(2 m/s2)(5 s)2
24
Step 3: Calculate the distance traveled by the car:
x=1
2(2)(25) = 25 m
Therefore, the car travels 25 meters in the first 5 seconds.
Question 31
Question
A car initially at rest starts accelerating at a constant rate of 3 m/s2for 10
seconds. After this time, the car undergoes a constant deceleration of 2 m/s2. If
the car comes to a stop after a total time of 30 seconds, what is the maximum
speed reached by the car during this time?
Solution
Step 1: Find the distance traveled during acceleration.
The distance traveled during acceleration is given by the equation:
d1=1
2at2
where ais the acceleration and tis the time.
Substitute a= 3 m/s2and t= 10 s:
d1=1
2×3×(10)2= 150 m
Step 2: Find the distance traveled during deceleration.
The deceleration will stop the car, so the distance traveled during decel-
eration is the remaining distance. We find that by subtracting the distances
already traveled from the total distance.
Total distance, dtotal =d1+d2, where d1= 150 m (from step 1) and a=
2 m/s2(deceleration).
Let d2be the distance during deceleration:
d2=dtotal −d1= 150 −1
2×2×(30 −10)2
Solve for d2:
d2= 150 −1
2×2×202= 50 m
Step 3: Find the maximum speed.
The maximum speed is reached when the car transitions from acceleration
to deceleration. At this point, the speed remains constant. We can find this
maximum speed using the formula:
speed = initial speed + a×t
25
The initial speed is 0 m/s, acceleration is 3 m/s
²
for 10 seconds, so the speed
at the transition point is:
speed = 0 + 3 ×10 = 30 m/s
Therefore, the maximum speed reached by the car is 30 m/s .
Question 32
Question
A particle moves along a straight line such that its acceleration is given by
a(t)=6t−2 m/s2. If the particle is at rest at t= 1 s and its initial velocity is
3 m/s, determine the particle’s velocity at t= 3 s.
Solution
Step 1: To find the velocity function v(t), we will integrate the given acceleration
function a(t) with respect to time.
Za(t)dt =Z(6t−2) dt
Za(t)dt = 3t2−2t+C
where Cis the constant of integration.
Step 2: We know that the particle is at rest at t= 1 s, so at t= 1 s, v(1) = 0
m/s.
0 = 3(1)2−2(1) + C
0=3−2 + C
C=−1
Step 3: Therefore, the velocity function v(t) is given by:
v(t)=3t2−2t−1
Step 4: Now, to find the velocity at t= 3 s, we substitute t= 3 into the
velocity function v(t).
v(3) = 3(3)2−2(3) −1
v(3) = 27 −6−1
v(3) = 20 m/s
So, the particle’s velocity at t= 3 s is 20 m/s.
26
Question 33
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2for 10 seconds. How
far does the car travel during this time?
Solution
Step 1: We are given the initial velocity (u= 0 m/s), acceleration (a= 3.0 m/s2),
and time interval (t= 10 s). We need to find the distance traveled by the car
during this time. Since the car starts from rest, its initial velocity is 0 m/s.
Step 2: We can use the kinematic equation for displacement in terms of
initial velocity, acceleration, and time:
s=ut +1
2at2
Step 3: Substituting the given values into the equation, we have
s= 0 ×10 + 1
2×3.0×(10)2
Step 4: Simplifying the expression, we get
s= 0 + 1
2×3.0×100
Step 5: Therefore, the distance traveled by the car during this time is
s=1
2×3.0×100 = 150 m
Step 6: Hence, the car travels a distance of 150 meters during the 10-second
acceleration period.
Question 34
Question
A car initially at rest starts moving along a straight road with a constant accel-
eration of 2 m/s2. After 5 seconds, what is the car’s velocity and displacement
from its initial position?
Solution
Step 1: First, we find the car’s velocity after 5 seconds using the equation of
motion:
v=u+at
27
where vis the final velocity, uis the initial velocity, ais the acceleration, and t
is the time. Substitute u= 0m/s, a= 2m/s2, and t= 5s into the equation:
v= 0 + 2 ×5 = 10 m/s
Step 2: Next, we find the displacement of the car after 5 seconds using the
equation of motion:
s=ut +1
2at2
where sis the displacement. Substitute u= 0m/s, a= 2m/s2, and t= 5s into
the equation:
s= 0 ×5 + 1
2×2×52= 0 + 0 + 25 = 25 m
Therefore, after 5 seconds, the car’s velocity is 10 m/s and its displacement
from its initial position is 25 meters.
Question 35
Question
An object moves along a straight line with an initial velocity of 5 m/s and a
constant acceleration of -2 m/s2. What is the distance traveled by the object
when it first comes to rest?
Solution
Let’s denote the initial velocity as v0= 5 m/s, the acceleration as a=−2 m/s2,
the final velocity as vf= 0 m/s, and the distance traveled as d.
Step 1: Use the kinematic equation v2
f=v2
0+ 2ad to find the distance
traveled. Since vf= 0, we have:
0 = (5)2+ 2(−2)d
Step 2: Solve for d:
25 = −4d
d=−25
4
d=−6.25 m
Therefore, the object travels a distance of 6.25 meters before coming to rest.
28
