PHYS 202 - GENERAL PHYSICS II -
Kinematics in One Dimension
Question Bank - Set 5
Liberty University
Question 1
Question
A car starts from rest and accelerates uniformly at 2 m/s2for 10 seconds. Cal-
culate the final velocity of the car at the end of this time period.
Solution
Step 1: Determine the acceleration of the car. Given that the car accelerates
uniformly at 2 m/s2, the acceleration (a) is 2 m/s2.
Step 2: Determine the initial velocity of the car. Since the car starts from
rest, the initial velocity (u) is 0 m/s.
Step 3: Use the kinematic equation v=u+at to find the final velocity (v)
of the car. Plug in the values of u,a, and tinto the equation:
v= 0 + (2 m/s2)(10 s)
v= 0 + 20
v= 20 m/s
Therefore, the final velocity of the car at the end of 10 seconds is 20 m/s.
Question 2
Question
A car starts from rest and accelerates uniformly to a speed of 25 m/s in a time
of 10 seconds. What is the acceleration of the car?
Solution
Step 1: Identify the known variables: The final velocity of the car, vf, is 25
m/s. The initial velocity of the car, vi, is 0 m/s (as the car starts from rest).
The time taken for the car to reach its final velocity, t, is 10 seconds.
Step 2: Use the kinematic equation to relate the variables: We’ll use the
equation vf=vi+a·t, where ais the acceleration.
Step 3: Plug in the known values and solve for acceleration: Substitute
vf= 25 m/s, vi= 0 m/s, and t= 10 s into the equation: 25 = 0 + a·10
Step 4: Solve for acceleration: a=25
10 = 2.5 m/s2
Therefore, the acceleration of the car is 2.5 m/s2.
Question 3
Question
A car accelerates uniformly from rest for 10 seconds along a straight road. After
this time, the car reaches a speed of 30 m/s. Find the acceleration of the car
during this time period.
Solution
Step 1: Identify the given variables. Let’s denote the initial velocity of the car
as vi(which is 0 m/s since the car starts from rest), the final velocity as vf
(which is 30 m/s), the acceleration as a, and the time taken as t(which is 10
seconds).
Step 2: Use the kinematic equation vf=vi+a·tto find the acceleration.
Substitute the given values into the equation:
30 m/s = 0 m/s + a·10 s
Solve for acceleration a:
a=30 m/s
10 s = 3 m/s2
Step 3: Verify the units and interpret the result. The acceleration of the car
during this time period is 3 m/s2. This means that each second, the car’s speed
increases by 3 m/s.
Question 4
Question
A car starts from rest and accelerates at 2.5 m/s2for 8 seconds.
1. What is the final velocity of the car at the end of the 8 seconds?
2. How far has the car traveled during this time?
2
Solution
1. To find the final velocity of the car, we can use the kinematic equation:
v=u+at
where: - vis the final velocity, - uis the initial velocity (which is 0 because the
car starts from rest), - ais the acceleration, - tis the time.
Step 1: Calculate the final velocity:
v= 0 + (2.5 m/s2)(8 s)
v= 20 m/s
Therefore, the final velocity of the car at the end of 8 seconds is 20 m/s.
2. To find the distance traveled by the car during this time, we can use the
kinematic equation for distance:
s=ut +1
2at2
where: - sis the distance, - uis the initial velocity, - ais the acceleration, - tis
the time.
Since the car starts from rest, the initial velocity u= 0.
Step 1: Calculate the distance traveled:
s= 0 + 1
2(2.5 m/s2)(8 s)2
s= 80 m
Therefore, the car has traveled 80 meters during this time.
Question 5
Question
A car accelerates from rest at a constant rate of 2 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 at the end of the 10-second
interval using the kinematic equation:
v=u+at
where: v= final velocity = ? u= initial velocity = 0 m/s (rest) a= acceleration
= 2 m/s2t= time interval = 10 s
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Substitute the values into the equation:
v= 0 + (2)(10) = 20 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
where: s= distance traveled = ? u= initial velocity = 0 m/s a= acceleration
= 2 m/s2t= time interval = 10 s
Substitute the values into the equation:
s= (0)(10) + 1
2(2)(10)2
s= 0 + 1
2(2)(100) = 100 m
Therefore, the total distance traveled by the car during the 10-second interval
is 100 meters.
Question 6
Question
An object starts from rest and accelerates uniformly at 2 m/s2. After 5 seconds,
the object encounters a force that causes it to decelerate uniformly at 1 m/s2.
If the object comes to a stop after a total of 10 seconds, determine the total
distance traveled by the object.
Solution
Step 1: Determine the distance covered during the acceleration phase. Given
that the object starts from rest and accelerates uniformly at 2 m/s2for 5 seconds,
we can use the kinematic equation:
d=vit+1
2at2
where: - dis the displacement, - viis the initial velocity (which is 0 m/s since
the object starts from rest), - ais the acceleration, and - tis the time.
Plugging in the values:
d= (0 m/s)(5 s) + 1
2(2 m/s2)(5 s)2
d= 0 + 1
2(2)(25)
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d= 25 m
Therefore, the distance covered during the acceleration phase is 25 meters.
Step 2: Determine the distance covered during the deceleration phase. Given
that the object decelerates uniformly at 1 m/s2for the remaining 5 seconds, we
can use the same kinematic equation:
d=vit+1
2at2
Since the object comes to a stop at the end of deceleration phase, the final
velocity is 0 m/s. Plugging in the values:
d= (0 m/s)(5 s) + 1
2(−1 m/s2)(5 s)2
d=−1
2(25)
d=−12.5 m
Therefore, the distance covered during the deceleration phase is -12.5 meters
(negative because it’s in the opposite direction).
Step 3: Calculate the total distance traveled by the object. The total dis-
tance traveled by the object is the sum of the distances covered during the
acceleration and deceleration phases:
Total distance = 25 m + (−12.5 m)
Total distance = 12.5 m
Therefore, the total distance traveled by the object is 12.5 meters.
Question 7
Question
A car traveling at 30 m/s slows down with a constant acceleration of -2.0 m/s2.
What is the car’s velocity after 5 seconds?
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, vi= 30 m/s
Acceleration, a=−2.0 m/s2
Time, t= 5 s
Unknown: Final velocity, vf.
Step 2: Use the kinematic equation vf=vi+a·tto solve for the final
velocity.
Substitute the given values into the equation:
vf= 30 m/s −2.0 m/s2·5 s
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Step 3: Calculate the final velocity.
vf= 30 m/s −2.0 m/s2·5 s = 30 m/s −10 m/s = 20 m/s
Therefore, the final velocity of the car after 5 seconds is 20 m/s.
Question 8
Question
A car accelerates from rest at a constant rate of 3 m/s2for 8 seconds. After
that, the car maintains a constant velocity for 10 seconds, and then decelerates
at a rate of 2 m/s2until it comes to a stop. What is the total distance traveled
by the car during this entire motion?
Solution
Step 1: Find the distance traveled during acceleration. Given that the car
accelerates from rest at a rate of 3 m/s2for 8 seconds, we can use the kinematic
equation:
d=1
2at2
where dis the distance traveled, ais the acceleration, and tis the time. Sub-
stitute a= 3 m/s2and t= 8 s:
d=1
2×3×(8)2
d= 96 m
Step 2: Find the distance traveled at constant velocity. During the time
where the car maintains a constant velocity, the distance traveled is given by:
distance = velocity ×time
Since the car maintains a constant velocity, the distance traveled at constant
velocity is:
distance = velocity ×time = 3 m/s ×10 s = 30 m
Step 3: Find the distance traveled during deceleration. The deceleration
rate is 2 m/s2until the car comes to a stop. The distance traveled during
deceleration can be calculated using the equation:
d=v2
f−v2
i
2a
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where vfis the final velocity, viis the initial velocity, and ais the deceleration
rate. The initial velocity during deceleration is the constant velocity of 3 m/s.
The final velocity is 0 m/s when the car comes to a stop.
d=(0)2−(3)2
2(−2)
d=−9
−4=9
4= 2.25 m
Step 4: Calculate the total distance traveled. The total distance traveled by
the car is the sum of the distances during acceleration, at constant velocity, and
during deceleration.
Total distance = 96 m + 30 m + 2.25 m = 128.25 m
Therefore, the total distance traveled by the car during this entire motion is
128.25 meters.
Question 9
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2for 10
seconds along a straight road. After this time, the car continues at a constant
speed for an additional 20 seconds. What is the total distance traveled by the
car during this time interval?
Solution
Step 1: Find the distance traveled during the acceleration phase. Given that
the car starts from rest and accelerates at a constant rate of 2.5 m/s2for 10
seconds, we can find the distance traveled during this time using the equation
for motion:
d1=1
2·a·t2
where: a= acceleration = 2.5 m/s2(constant) t= time = 10 seconds
Substitute the values into the equation:
d1=1
2·2.5·(10)2
d1= 0.5·2.5·100
d1= 125 m
Therefore, the distance traveled during the acceleration phase is 125 meters.
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Step 2: Find the distance traveled during the constant speed phase. During
the constant speed phase, the car continues for an additional 20 seconds. Since
the speed is constant, we can find the distance traveled using the equation:
d2=v·t
where: v= constant speed
Since the car continues at a constant speed, the distance traveled during this
phase is:
d2=v·20
Step 3: Find the total distance traveled by the car. The total distance
traveled is the sum of the distances traveled during the acceleration and constant
speed phases:
Total distance = d1+d2= 125 + (v·20)
To find the constant speed v, we use the fact that the car started from rest:
v=a·t= 2.5·10 = 25 m/s
Therefore,
Total distance = 125 + (25 ·20) = 125 + 500 = 625 m
Therefore, the total distance traveled by the car during this time interval is
625 meters.
Question 10
Question
A car traveling at a constant velocity of 25 m/s passes a stationary police car.
The police car starts accelerating at a rate of 2 m/s2just as the speeding car
passes. After what time interval and at what distance will the police car catch
up to the speeding car?
Solution
Step 1: Determine the time it takes for the police car to catch up to the speeding
car. Let tbe the time it takes for the police car to catch up to the speeding car.
The position of the speeding car can be described by the equation:
xspeeding = 25t
The position of the police car can be described by the equation:
xpolice =1
2·2t2
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Step 2: Set up an equation to find the time at which the police car catches
up to the speeding car. The police car catches up to the speeding car when the
positions of both cars are equal:
25t=1
2·2t2
Step 3: Solve for t. This equation simplifies to:
25t= 2t2
2t2−25t= 0
t(2t−25) = 0
This equation has two possible solutions: t= 0 and t= 12.5.
Step 4: Analyze the solutions. The solution t= 0 represents the initial
time when the police car starts accelerating and the speeding car passes. The
solution t= 12.5 represents the time it takes for the police car to catch up to
the speeding car.
Step 5: Find the distance at which the police car catches up to the speeding
car.
xspeeding = 25 ×12.5 = 312.5 m
Therefore, the police car catches up to the speeding car after 12.5 seconds
and at a distance of 312.5 meters from the starting point.
Question 11
Question
A car is moving along a straight road. The acceleration of the car is given by
a(t)=2t−1 m/s2, where tis the time in seconds and t≥0. If the car starts
from rest at t= 0, determine the car’s velocity and position as a function of
time.
Solution
Step 1: To find the velocity function v(t), we need to integrate the acceleration
function a(t).
a(t) = dv
dt = 2t−1
Integrating both sides with respect to t:
Zdv =Z(2t−1)dt
v(t) = Z(2t−1)dt
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v(t) = t2−t+C1
Step 2: Since the car starts from rest, we know that v(0) = 0.
0=02−0 + C1
C1= 0
So, the velocity function is given by:
v(t) = t2−t
Step 3: To find the position function x(t), we need to integrate the velocity
function v(t).
v(t) = dx
dt =t2−t
Integrating both sides with respect to t:
Zdx =Z(t2−t)dt
x(t) = Z(t2−t)dt
x(t) = 1
3t3−1
2t2+C2
Step 4: Since the car starts from rest, we know that x(0) = 0.
0 = 1
3·03−1
2·02+C2
C2= 0
So, the position function is given by:
x(t) = 1
3t3−1
2t2
Question 12
Question
A car accelerates from rest with a constant acceleration of 4 m/s2. At the same
time, a truck 240 m ahead of the car starts moving with a constant velocity of
10 m/s. How far from the truck does the car overtake it?
10
Solution
Let’s denote the initial position of the car and the truck as xcar(0) = 0 m and
xtruck(0) = 240 m, respectively. The initial velocities are vcar(0) = 0 m/s and
vtruck(0) = 10 m/s, respectively. The acceleration of the car is acar = 4 m/s2.
The equations to find the position of the car and truck as functions of time
are:
For the car:
xcar(t) = 1
2acart2
For the truck:
xtruck(t) = vtruckt+xtruck(0)
To find the time it takes for the car to overtake the truck, we set xcar(t) =
xtruck(t):
1
2acart2=vtruckt+xtruck(0)
Solving for t, we get:
2t2−10t−240 = 0
This is a quadratic equation, whose solutions are t= 12 s (ignoring the
negative solution).
Now, let’s find how far from the truck the car overtakes it:
xovertake =xcar(t)
Step 1: Calculate the position of the car at time t= 12 s.
xcar(12) = 1
2×4×(12)2= 288 m
Therefore, the car overtakes the truck 288 m from its initial position.
Question 13
Question
A car starts from rest at a stop sign and accelerates uniformly at a rate of 2.5
m/s2in a straight line for a distance of 150 m. What is the final velocity of the
car?
Solution
Step 1: Identify the known variables. The initial velocity (v0) of the car is 0
m/s. The acceleration (a) of the car is 2.5 m/s2. The distance (d) the car travels
is 150 m. The final velocity is what we’re trying to find (vf).
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Step 2: Choose the appropriate kinematic equation to solve for the final
velocity. Since we are not given the time taken for the car to accelerate, we can
use the kinematic equation:
v2
f=v2
0+ 2ad
Step 3: Plug in the known values into the kinematic equation and solve for
the final velocity. Substitute the values into the formula:
v2
f= (0)2+ 2(2.5)(150)
v2
f= 0 + 750
vf=√750
vf≈27.39 m/s
Step 4: State the final velocity of the car. Therefore, the final velocity of
the car is approximately 27.39 m/s.
Question 14
Question
A car starts from rest at t= 0 and accelerates uniformly at 2 m/s2for 5 sec-
onds. It then maintains a constant velocity for 10 seconds before decelerating
uniformly at 1 m/s2until it comes to a stop. What is the total distance the car
travels during this entire motion?
Solution
Step 1: Find the distance traveled during acceleration from rest at t= 0 to
t= 5 seconds. The distance traveled during acceleration can be found using the
formula: x=1
2at2, where ais the acceleration, and tis the time. Substitute
a= 2 m/s2and t= 5 seconds: x=1
2×2 m/s2×(5 s)2= 25 m.
Step 2: Find the distance traveled during constant velocity from t= 5 to
t= 15 seconds. Since the velocity is constant during this time interval, the
distance traveled is simply the product of velocity and time. The velocity after
5 seconds is 2 m/s ×5 s = 10 m.
Step 3: Find the distance traveled during deceleration from t= 15 to t= 25
seconds. The distance traveled during deceleration can be found using the
formula: x=vit+1
2at2, where viis the initial velocity, ais the acceleration,
and tis the time. The final velocity is 0 since the car comes to a stop. Substitute
vi= 10 m/s, a=−1 m/s2(negative due to deceleration), and t= 10 seconds:
x= 10 m/s ×10 s + 1
2×(−1 m/s2)×(10 s)2= 50 m.
Step 4: Calculate the total distance traveled by summing the distances from
each step. Total distance = 25 m + 10 m + 50 m = 85 m.
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Question 15
Question
A car accelerates from rest at a constant rate of 2 m/s2for 10 seconds. After
this time, the car maintains a constant speed for 20 seconds before decelerating
to a stop at a rate of 3 m/s2. What is the total distance traveled by the car
during this entire trip?
Solution
Step 1: Find the distance traveled during acceleration. The distance traveled
during acceleration can be found using the equation:
d=1
2at2
where ais the acceleration and tis the time. Substitute a= 2 m/s2and t= 10 s:
d=1
2×2×(10)2
d=1
2×2×100
d= 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 velocity and tis the time. Since the car maintains a constant
speed during this time, vis the same as the final velocity after acceleration. The
final velocity after acceleration is:
v=a×t= 2 ×10 = 20 m/s
Substitute v= 20 m/s and t= 20 s:
d= 20 ×20
d= 400 m
Step 3: Find the distance traveled during deceleration. The distance traveled
during deceleration can also be found using the equation:
d=1
2at2
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where ais the deceleration and tis the time. Substitute a=−3 m/s2(since it’s
deceleration) and t= 20 s:
d=1
2× −3×(20)2
d=1
2× −3×400
d=−600 m
Step 4: Calculate the total distance traveled. The total distance traveled
is the sum of the distances traveled during acceleration, constant velocity, and
deceleration:
Total distance = 100 m + 400 m −600 m
Total distance = 100 m + 400 m −600 m
Total distance = −100 m
Therefore, the total distance traveled by the car during the entire trip is 100
meters.
Question 16
Question
A car accelerates uniformly from rest at a rate of 3.00 m/s2for 8.00 seconds,
then maintains a constant velocity for 15.00 seconds, and finally decelerates
uniformly to a stop in 13.00 seconds. What is the average velocity of the car
during the entire 36.00-second trip?
Solution
Step 1: Find the distance traveled during the acceleration phase. Using the
equation for position at time tduring constant acceleration:
s=vit+1
2at2
where sis the distance traveled, viis the initial velocity, ais the acceleration,
and tis the time. Given: vi= 0 m/s, a= 3.00 m/s2,t= 8.00 s. Plugging in the
values:
s= 0 + 1
2×3.00 ×(8.00)2= 96.00 m
Step 2: Find the distance traveled during the deceleration phase. Given:
vf= 0 m/s, a=−3.00 m/s2,t= 13.00 s. Using the same formula and plugging
in the values:
s= 0 + 1
2× −3.00 ×(13.00)2=−253.50 m
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The negative sign indicates opposite direction of motion.
Step 3: Find the average velocity during the entire trip. The total distance
covered by the car is the sum of the distances covered during acceleration,
constant velocity, and deceleration.
dtotal = 96.00 m + 0 m + (−253.50 m) = −157.50 m
The average velocity ¯vis given by:
¯v=dtotal
ttotal
where ttotal = 36.00 s. Plugging in the values:
¯v=−157.50 m
36.00 s =−4.38 m/s
Therefore, the average velocity of the car during the entire 36.00-second trip
is −4.38 m/s.
Question 17
Question
A car starts from rest and accelerates at a constant rate of 3 m/s2for a distance
of 100 meters. Calculate the final velocity of the car.
Solution
Step 1: We can use the kinematic equation relating final velocity, initial velocity,
acceleration, and displacement:
v2=u2+ 2as
where - vis the final velocity, - uis the initial velocity (in this case, the car
starts from rest so u= 0), - ais the acceleration, and - sis the displacement.
Step 2: Substituting the given values into the equation, we have:
v2= 0 + 2(3)(100)
v2= 600
Step 3: Taking the square root of both sides to find the final velocity:
v=√600
v≈24.49 m/s
Therefore, the final velocity of the car is approximately 24.49 m/s.
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Question 18
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2for 10 seconds.
Then the car moves with a constant velocity for the next 20 seconds. If the car
finally decelerates uniformly at 1.0 m/s2until it stops, find the total distance
traveled by the car during the entire motion.
Solution
Step 1: Calculate the distance traveled during acceleration phase. The dis-
tance traveled during acceleration phase can be calculated using the kinematic
equation:
d=vit+1
2at2
where: - viis the initial velocity (0 m/s), - ais the acceleration (2.0 m/s2), - t
is the time (10 seconds).
Substitute the values into the equation:
d= 0 + 1
2×2.0×(10)2= 100 m
So, the car travels 100 meters during the acceleration phase.
Step 2: Calculate the distance traveled during constant velocity phase. Since
the car moves with a constant velocity, the distance traveled is simply the prod-
uct of the velocity and time:
d=vt
where: - vis the constant velocity, - tis the time (20 seconds).
Since velocity is distance divided by time, we can rearrange the equation to
find the distance:
d=dconstant velocity
20 ×20 = dconstant velocity
Thus, the distance traveled during the constant velocity phase is dependent on
the velocity during that phase.
Step 3: Calculate the distance traveled during deceleration phase. The dis-
tance traveled during deceleration phase can be calculated using the same kine-
matic equation:
d=vit+1
2at2
where: - viis the initial velocity (the velocity after the constant velocity phase),
-ais the deceleration (-1.0 m/s2), - tis the time (unknown).
The velocity after the constant velocity phase is the same as the constant
velocity during that phase. This velocity can be calculated using the fact that
acceleration is the rate of change of velocity.
a=vf−vi
t
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Since the car moves with constant velocity, the final velocity after 20 seconds is
the same as the constant velocity. Thus, vf=vconstant velocity. We can rearrange
the equation to find vconstant velocity in terms of acceleration:
2.0 = vconstant velocity −0
10
vconstant velocity = 20 m/s
Substitute the values into the kinematic equation:
d= 20 ×10 + 1
2×(−1.0) ×t2
20t=1
2t2
t= 40 seconds
Therefore, the distance traveled during deceleration phase is:
d= 20 ×40 + 1
2×(−1.0) ×(40)2= 800 m
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of distances traveled during each phase:
Total distance = 100 + dconstant velocity + 800
Total distance = 100 + 20 ×20 + 800 = 900 meters
Thus, the total distance traveled by the car during the entire motion is 900
meters.
Question 19
Question
A car starts from rest and accelerates along a straight road with a constant
acceleration of 3.0 m/s2. At the same instant, a truck passes the car from
behind moving with a constant velocity of 20.0 m/s.
1. How far does the car travel before it overtakes the truck?
2. How long does it take for the car to overtake the truck?
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Solution
Let’s denote the initial position of the car and the truck as 0 meters. We will
first find the time it takes for the car to overtake the truck, and then use this
time to find the distance the car has traveled at that time.
Step 1: Find the time for the car to overtake the truck. The position
of the car as a function of time can be described by the equation:
xcar(t) = 1
2at2
Where: - a= 3.0 m/s2is the acceleration of the car. - xcar(t) is the position of
the car at time t.
The position of the truck as a function of time can be described by the
equation:
xtruck(t) = 20t
Where: - xtruck(t) is the position of the truck at time t.
The car overtakes the truck when their positions are the same:
xcar(t) = xtruck(t)
1
2·3.0t2= 20t
1.5t2= 20t
1.5t= 20
t=20
1.5= 13.33 s
Step 2: Find the distance the car travels before overtaking the
truck. Using the time found in Step 1, we can find the distance the car has
traveled at that time:
xcar(13.33) = 1
2·3.0·(13.33)2
xcar(13.33) = 89.21 m
Thus, 1. The car travels 89.21 meters before overtaking the truck. 2. It
takes 13.33 seconds for the car to overtake the truck.
Question 20
Question
A car accelerates uniformly from rest and reaches a velocity of 25 m/s in 10
seconds. Calculate the distance traveled by the car during this time.
18
Solution
Step 1: Find the acceleration of the car using the formula a=vf−vi
t, where a
is the acceleration, vfis the final velocity, viis the initial velocity, and tis the
time.
Given: vf= 25 m/s, vi= 0 m/s, t = 10 s
Acceleration: a=25 m/s −0 m/s
10 s = 2.5 m/s2
Step 2: Calculate the distance traveled by the car using the formula d=
vit+1
2at2, where dis the distance traveled, viis the initial velocity, tis the
time, and ais the acceleration.
Distance: d= 0 ×10 + 1
2×2.5×(10)2
d= 0 + 1
2×2.5×100
d=1
2×250 = 125 m
Therefore, the car travels a distance of 125 meters during the 10-second
intervals.
Question 21
Question
A car is initially at rest at t= 0 and accelerates uniformly at 2 m/s2. At t= 5
s, the velocity of the car is measured to be 10 m/s. What is the position of the
car at t= 10 s?
Solution
Step 1: Determine the acceleration of the car using the given information.
The acceleration of the car is given as 2 m/s2.
Step 2: Find the velocity of the car at t= 10 s.
Using the equation for uniformly accelerated motion:
v=u+at
where: - vis the final velocity (10 m/s), - uis the initial velocity (0 m/s), - ais
the acceleration (2 m/s2), - tis the time interval (5 s).
Substitute the known values into the formula:
10 = 0 + 2 ×5
10 = 10 m/s
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Step 3: Calculate the displacement of the car at t= 10 s.
To find the displacement, we use the equation for displacement in uniformly
accelerated motion:
s=ut +1
2at2
where: - sis the displacement, - uis the initial velocity, - ais the acceleration,
-tis the time interval.
Substitute the values:
s= 0 ×10 + 1
2×2×102
s= 0 + 100
s= 100 m
Therefore, the position of the car at t= 10 s is 100 meters.
Question 22
Question
A car accelerates from rest at a constant rate of 4 m/s2. How far does the car
travel in the first 5 seconds?
Solution
Step 1: Determine the acceleration of the car
Given that the car accelerates at a constant rate of 4 m/s2, we have a= 4 m/s2.
Step 2: Determine the initial velocity of the car
Since the car starts from rest, the initial velocity v0is 0 m/s.
Step 3: Use the kinematic equation
The kinematic equation we can use to find the distance traveled by the car is:
x=v0t+1
2at2
Step 4: Plug in the values and solve for x
Substitute v0= 0 m/s, a= 4 m/s2, and t= 5 s into the equation:
x= 0 ×5 + 1
2×4×(5)2
x= 0 + 10 ×5
x= 50 m
Therefore, the car travels 50 meters in the first 5 seconds.
20
Question 23
Question
A car starts from rest at t= 0 and accelerates at a constant rate of 2.0 m/s2.
At what time after it started does the car’s speed reach 25 m/s?
Solution
Step 1: Let’s denote the final speed of the car as v, acceleration as a, initial
speed as u(which is 0 since the car starts from rest), and time as t. We are
given a= 2.0 m/s2and v= 25 m/s.
Step 2: We can use the kinematic equation v=u+at, where vis the final
speed, uis the initial speed, ais the acceleration, and tis the time.
Step 3: Plugging in the given values into the equation, we get:
25 = 0 + 2.0t
Step 4: Solving for t, we find:
t=25
2.0= 12.5 s
Step 5: Therefore, the car’s speed reaches 25 m/s at 12.5 seconds after it
started.
Question 24
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2for 8.0 seconds.
After this time, the car continues at a constant velocity for an additional 12
seconds, before coming to a stop with a uniform acceleration of -2.0 m/s2.
Determine the total distance traveled by the car during this entire motion.
Solution
Step 1: Calculate the distance traveled during the first acceleration phase. To
find the distance traveled during the acceleration phase, we can use the equation:
d=vit+1
2at2
where viis the initial velocity, tis the time, and ais the acceleration.
Given: Initial velocity, vi= 0 m/s (car starts from rest) Acceleration, a= 3.0
m/s2Time, t= 8.0 s
Plugging in the values, we get:
d= (0 ×8.0) + 1
2×3.0×(8.0)2
21
d= 0 + 1
2×3.0×64.0
d= 0 + 96.0
d= 96.0 m
Therefore, the distance traveled during the first acceleration phase is 96.0
m.
Step 2: Calculate the distance traveled during the constant velocity phase.
During the constant velocity phase, the distance traveled is given by:
d=vt
where vis the constant velocity and tis the time.
Given: Constant velocity during this phase, v= 3.0 m/s Time, t= 12.0 s
Plugging in the values, we get:
d= 3.0×12.0
d= 36.0 m
Therefore, the distance traveled during the constant velocity phase is 36.0
m.
Step 3: Calculate the distance traveled during the deceleration phase. To
find the distance traveled during the deceleration phase, we can use the same
equation as in Step 1, but with the negative acceleration value:
d=vit+1
2at2
where viis the initial velocity (which is the constant velocity in this case), tis
the time, and ais the deceleration.
Given: Initial velocity (constant velocity), vi= 3.0 m/s Deceleration, a=
−2.0 m/s2Time, t= 8.0 s
Plugging in the values, we get:
d= 3.0×8.0 + 1
2× −2.0×(8.0)2
d= 24.0 + 1
2× −2.0×64.0
d= 24.0−64.0
d=−40.0 m
Therefore, the distance traveled during the deceleration phase is 40.0 m
(negative sign indicates the direction of travel).
Step 4: Calculate the total distance traveled by the car. The total distance
traveled by the car is the sum of the distances calculated in Step 1, Step 2, and
Step 3: Total distance = Distance in acceleration phase + Distance in constant
velocity phase + Distance in deceleration phase Total distance = 96.0 + 36.0 -
40.0 Total distance = 92.0 m
Therefore, the total distance traveled by the car during this entire motion is
92.0 meters.
22
Question 25
Question
A particle moves along the x-axis according to the equation x(t)=4t3−3t2+ 2,
where xis in meters and tis in seconds. Determine the velocity and acceleration
of the particle when t= 2 s.
Solution
Step 1: Find the velocity of the particle by taking the derivative of the position
function with respect to time.
Given x(t) = 4t3−3t2+ 2
Velocity, v(t) = dx
dt =d
dt(4t3−3t2+ 2)
v(t) = 12t2−6t
Step 2: Find the velocity of the particle when t= 2 s.
v(2) = 12(2)2−6(2)
v(2) = 48 −12
v(2) = 36 m/s
Step 3: Find the acceleration of the particle by taking the derivative of the
velocity function with respect to time.
Acceleration, a(t) = dv
dt =d
dt(12t2−6t)
a(t) = 24t−6
Step 4: Find the acceleration of the particle when 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 36 m/s and the
acceleration is 42 m/s2.
Question 26
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2along a
straight road for 10 seconds. After this time, the car continues at a constant
speed for an additional 20 seconds. Find the total distance traveled by the car
during this time interval.
23
Solution
Step 1: Find the distance traveled during the acceleration phase. Since the car
starts from rest, the initial velocity, u= 0. The acceleration, a= 3.0 m/s2. The
time for acceleration, t= 10 s.
Using the equation s=ut +1
2at2, we can find the distance traveled during
the acceleration phase.
s= (0)(10) + 1
2(3.0)(10)2
s= 0 + 1
2(3.0)(100)
s= 150 m
Step 2: Find the distance traveled during the constant speed phase. During
the constant speed phase, the car’s velocity remains constant at the final velocity
reached at the end of the acceleration phase. The final velocity, v=u+at.
v= 0 + (3.0)(10)
v= 30 m/s
The distance traveled during the constant speed phase is given by s=vt.
s= (30)(20)
s= 600 m
Step 3: Find the total distance traveled. The total distance traveled is the
sum of the distances traveled during the acceleration and constant speed phases.
Total distance = 150 m + 600 m Total distance = 750 m
Question 27
Question
A car accelerates uniformly from rest to a speed of 25 m/s over a distance of
200 m. Find the time it takes for the car to reach this speed.
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, u= 0 m/s
Final velocity, v= 25 m/s
Distance traveled, s= 200 m
Unknown: Time taken to reach final speed, t
Step 2: Use the kinematic equation v=u+at where ais the acceleration.
Since the car starts from rest, u= 0, so the equation simplifies to v=at.
24
Step 3: Calculate the acceleration using the formula a=v−u
t. Substituting
the known values, we get: a=25 m/s−0 m/s
t=25 m/s
t.
Step 4: Use the kinematic equation s=ut +1
2at2. Since u= 0, the equation
simplifies to s=1
2at2.
Step 5: Substitute the values of sand ainto the equation. 200 m = 1
2×
25 m/s
t×t2=25t
2.
Step 6: Solve for tby rearranging the equation. 200 m = 25t
2
t=2×200
25 = 16 s.
Conclusion
It takes 16 seconds for the car to reach a speed of 25 m/s accelerating uniformly
over a distance of 200 m.
Question 28
Question
A race car starts from rest and accelerates uniformly to a speed of 100 m/s in
5 seconds. What is the average acceleration of the car during this time period?
Solution
Step 1: Identify the given values. The final velocity of the car, vf, is 100 m/s.
The initial velocity of the car, vi, is 0 m/s (since it starts from rest). The time
taken, t, is 5 seconds.
Step 2: Use the kinematic equation for uniform acceleration: vf=vi+at,
where ais the acceleration. Substitute the known values into the equation:
100 = 0 + a×5. Simplify the equation: 100 = 5a.
Step 3: Solve for the acceleration. Divide both sides by 5: a=100
5. There-
fore, a= 20 m/s2.
Step 4: Write the final answer. The average acceleration of the race car
during the 5-second time period is 20 m/s2.
Question 29
Question
A ball is thrown vertically upward from the ground with an initial speed of 30
m/s. Ignoring air resistance, determine the maximum height the ball reaches
above the ground.
25
Solution
Step 1: Write down the known variables and equations related to the motion
of the ball. Given: Initial velocity (u) = 30 m/s (upwards) Final velocity (v) at
maximum height = 0 m/s (when the ball momentarily stops) Acceleration (a)
=−9.81 m/s2(due to gravity, acting downwards)
The kinematic equation relating initial velocity, final velocity, acceleration,
and displacement is:
v2=u2+ 2as
Step 2: Determine the maximum height by rearranging the kinematic equa-
tion. At the maximum height, the final velocity is 0 m/s, so:
0 = (30)2+ 2(−9.81)s
0 = 900 −19.62s
19.62s= 900
s=900
19.62
Step 3: Calculate the maximum height.
s=900
19.62 ≈45.87 m
Therefore, the maximum height the ball reaches above the ground is approx-
imately 45.87 meters.
Question 30
Question
A car is initially at rest. It then accelerates uniformly along a straight road,
reaching a speed of 30 m/s in 10 seconds. What is the acceleration of the car?
Solution
Step 1: Identify the knowns
The initial velocity vi= 0 m/s, the final velocity vf= 30 m/s, and the time
t= 10 s.
Step 2: Use the kinematic equation
We can use the kinematic equation for uniformly accelerated motion:
vf=vi+at
Step 3: Substitute the known values
Substitute vi= 0 m/s, vf= 30 m/s, and t= 10 s into the equation:
30 = 0 + a×10
26
Step 4: Solve for acceleration
Solving for acceleration agives:
a=30
10 = 3 m/s2
Step 5: State the final answer
The acceleration of the car is 3 m/s2.
Question 31
Question
A car accelerates at a constant rate from rest, covering a distance of 400 m in
20 seconds. Find the speed of the car at the end of the 20 seconds.
Solution
Step 1: Identify the given quantities and the unknown. Let’s denote: - Initial
velocity of the car, vi= 0 m/s (starting from rest) - Distance covered, d= 400
m - Time taken, t= 20 s - Final velocity of the car, vf=?
Step 2: Use the kinematic equation d=vit+1
2at2to relate the given quan-
tities. The equation can be rearranged to solve for the final velocity vf:
vf=d−vit
t
Step 3: Substitute the given values into the equation.
vf=400 −0×20
20
vf=400
20
vf= 20 m/s
Answer: The speed of the car at the end of 20 seconds is 20 m/s.
Question 32
Question
A car starts from rest and accelerates uniformly to a speed of 30 m/s in 5
seconds. What is the acceleration of the car?
27
Solution
Step 1: Identify the given values and the unknown in the problem. The initial
velocity of the car, vi, is 0 m/s (starts from rest). The final velocity of the car,
vf, is 30 m/s. The time taken for the car to reach the final velocity, t, is 5
seconds. The acceleration of the car, a, is the unknown in the problem.
Step 2: Use the kinematic equation relating initial velocity, final velocity,
acceleration, and time. The kinematic equation we can use is:
vf=vi+a·t
Step 3: Substitute the known values into the kinematic equation. Substitute
vf= 30 m/s, vi= 0 m/s, and t= 5 s into the equation:
30 = 0 + a·5
Step 4: Solve for the acceleration. Simplify the equation to solve for accel-
eration:
30 = 5a
a=30
5
a= 6 m/s2
Step 5: State the final answer. The acceleration of the car is 6 m/s2.
Question 33
Question
A car travels along a straight road. Its velocity as a function of time is given
by v(t) = 6t2−4t+ 8, where vis in m/s and tis in seconds. Determine the
acceleration of the car as a function of time and find its displacement during
the time interval 1 ≤t≤3 seconds.
Solution
Step 1: Acceleration is the time derivative of velocity, so we need to find dv
dt .
Step 1: a(t) = dv
dt =d(6t2−4t+ 8)
dt = 12t−4
Step 2: To find the displacement, we need to integrate the velocity function
over the given time interval. Using the second fundamental theorem of calculus,
Step 2: Displacement = Z3
1
v(t)dt =Z3
1
(6t2−4t+ 8)dt
=2t3−2t2+ 8t3
1= (2(3)3−2(3)2+ 8(3)) −(2(1)3−2(1)2+ 8(1))
28
= (54 −18 + 24) −(2 −2 + 8) = 60 −8 = 52 meters
Therefore, the acceleration of the car as a function of time is a(t) = 12t−4
m/s2and its displacement during the time interval 1 ≤t≤3 seconds is 52
meters.
Question 34
Question
A car is traveling with a velocity given by v(t) = 10 −2tm/s, where tis the
time in seconds. Find the acceleration of the car as a function of time.
Solution
To find the acceleration of the car as a function of time, we need to differentiate
the velocity function with respect to time t.
Step 1: Determine the expression for acceleration a(t) using the given ve-
locity function.
v(t) = 10 −2t
Step 2: Differentiate the velocity function with respect to time to find the
acceleration.
a(t) = dv
dt
a(t) = d(10 −2t)
dt
a(t) = −2
Step 3: Write down the final expression for the acceleration of the car as a
function of time. So, the acceleration of the car is a constant -2 m/s2.
Question 35
Question
A car is traveling on a straight road. The car starts from rest and accelerates
uniformly at 2.0 m/s2for 10 seconds. It then maintains a constant speed for
20 seconds before decelerating uniformly at 1.5 m/s2until it comes to a stop.
What is the total distance traveled by the car during this entire process?
29
Solution
Step 1: Find the distance traveled during the acceleration phase. The formula
for distance traveled during uniform acceleration is given by:
d=1
2at2
where dis the distance traveled, ais the acceleration, and tis the time.
Given that a= 2.0m/s2and t= 10 s, we can substitute these values into
the formula to find the distance d1traveled during acceleration phase.
d1=1
2×2.0×(10)2
d1=1
2×2.0×100
d1= 100 m
Therefore, the distance traveled during the acceleration phase is 100 meters.
Step 2: Find the distance traveled during the constant speed phase. The
distance traveled during constant speed is given by:
d=vt
where dis the distance traveled, vis the constant speed, and tis the time.
Given that the car maintains a constant speed, the distance traveled during
this phase is:
d2=v×20
Since the speed is constant, the distance traveled during this phase is simply
the speed multiplied by time.
Step 3: Find the distance traveled during the deceleration phase. Similar
to the acceleration phase, the distance traveled during uniform deceleration is
given by:
d=1
2at2
Given that the car decelerates at 1.5 m/s2until it comes to a stop, we can
find the distance d3traveled during the deceleration phase.
d3=1
2×1.5×t2
Step 4: Calculate the total distance traveled by adding the distances from
each phase. The total distance traveled by the car is:
Total distance = d1+d2+d3
30
Solution
Step 1: Identify the known variables: The final velocity of the car, vf, is 25
m/s. The initial velocity of the car, vi, is 0 m/s (as the car starts from rest).
The time taken for the car to reach its final velocity, t, is 10 seconds.
Step 2: Use the kinematic equation to relate the variables: We’ll use the
equation vf=vi+a·t, where ais the acceleration.
Step 3: Plug in the known values and solve for acceleration: Substitute
vf= 25 m/s, vi= 0 m/s, and t= 10 s into the equation: 25 = 0 + a·10
Step 4: Solve for acceleration: a=25
10 = 2.5 m/s2
Therefore, the acceleration of the car is 2.5 m/s2.
Question 3
Question
A car accelerates uniformly from rest for 10 seconds along a straight road. After
this time, the car reaches a speed of 30 m/s. Find the acceleration of the car
during this time period.
Solution
Step 1: Identify the given variables. Let’s denote the initial velocity of the car
as vi(which is 0 m/s since the car starts from rest), the final velocity as vf
(which is 30 m/s), the acceleration as a, and the time taken as t(which is 10
seconds).
Step 2: Use the kinematic equation vf=vi+a·tto find the acceleration.
Substitute the given values into the equation:
30 m/s = 0 m/s + a·10 s
Solve for acceleration a:
a=30 m/s
10 s = 3 m/s2
Step 3: Verify the units and interpret the result. The acceleration of the car
during this time period is 3 m/s2. This means that each second, the car’s speed
increases by 3 m/s.
Question 4
Question
A car starts from rest and accelerates at 2.5 m/s2for 8 seconds.
1. What is the final velocity of the car at the end of the 8 seconds?
2. How far has the car traveled during this time?
2
Solution
1. To find the final velocity of the car, we can use the kinematic equation:
v=u+at
where: - vis the final velocity, - uis the initial velocity (which is 0 because the
car starts from rest), - ais the acceleration, - tis the time.
Step 1: Calculate the final velocity:
v= 0 + (2.5 m/s2)(8 s)
v= 20 m/s
Therefore, the final velocity of the car at the end of 8 seconds is 20 m/s.
2. To find the distance traveled by the car during this time, we can use the
kinematic equation for distance:
s=ut +1
2at2
where: - sis the distance, - uis the initial velocity, - ais the acceleration, - tis
the time.
Since the car starts from rest, the initial velocity u= 0.
Step 1: Calculate the distance traveled:
s= 0 + 1
2(2.5 m/s2)(8 s)2
s= 80 m
Therefore, the car has traveled 80 meters during this time.
Question 5
Question
A car accelerates from rest at a constant rate of 2 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 at the end of the 10-second
interval using the kinematic equation:
v=u+at
where: v= final velocity = ? u= initial velocity = 0 m/s (rest) a= acceleration
= 2 m/s2t= time interval = 10 s
3
Substitute the values into the equation:
v= 0 + (2)(10) = 20 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
where: s= distance traveled = ? u= initial velocity = 0 m/s a= acceleration
= 2 m/s2t= time interval = 10 s
Substitute the values into the equation:
s= (0)(10) + 1
2(2)(10)2
s= 0 + 1
2(2)(100) = 100 m
Therefore, the total distance traveled by the car during the 10-second interval
is 100 meters.
Question 6
Question
An object starts from rest and accelerates uniformly at 2 m/s2. After 5 seconds,
the object encounters a force that causes it to decelerate uniformly at 1 m/s2.
If the object comes to a stop after a total of 10 seconds, determine the total
distance traveled by the object.
Solution
Step 1: Determine the distance covered during the acceleration phase. Given
that the object starts from rest and accelerates uniformly at 2 m/s2for 5 seconds,
we can use the kinematic equation:
d=vit+1
2at2
where: - dis the displacement, - viis the initial velocity (which is 0 m/s since
the object starts from rest), - ais the acceleration, and - tis the time.
Plugging in the values:
d= (0 m/s)(5 s) + 1
2(2 m/s2)(5 s)2
d= 0 + 1
2(2)(25)
4
d= 25 m
Therefore, the distance covered during the acceleration phase is 25 meters.
Step 2: Determine the distance covered during the deceleration phase. Given
that the object decelerates uniformly at 1 m/s2for the remaining 5 seconds, we
can use the same kinematic equation:
d=vit+1
2at2
Since the object comes to a stop at the end of deceleration phase, the final
velocity is 0 m/s. Plugging in the values:
d= (0 m/s)(5 s) + 1
2(−1 m/s2)(5 s)2
d=−1
2(25)
d=−12.5 m
Therefore, the distance covered during the deceleration phase is -12.5 meters
(negative because it’s in the opposite direction).
Step 3: Calculate the total distance traveled by the object. The total dis-
tance traveled by the object is the sum of the distances covered during the
acceleration and deceleration phases:
Total distance = 25 m + (−12.5 m)
Total distance = 12.5 m
Therefore, the total distance traveled by the object is 12.5 meters.
Question 7
Question
A car traveling at 30 m/s slows down with a constant acceleration of -2.0 m/s2.
What is the car’s velocity after 5 seconds?
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, vi= 30 m/s
Acceleration, a=−2.0 m/s2
Time, t= 5 s
Unknown: Final velocity, vf.
Step 2: Use the kinematic equation vf=vi+a·tto solve for the final
velocity.
Substitute the given values into the equation:
vf= 30 m/s −2.0 m/s2·5 s
5
Step 3: Calculate the final velocity.
vf= 30 m/s −2.0 m/s2·5 s = 30 m/s −10 m/s = 20 m/s
Therefore, the final velocity of the car after 5 seconds is 20 m/s.
Question 8
Question
A car accelerates from rest at a constant rate of 3 m/s2for 8 seconds. After
that, the car maintains a constant velocity for 10 seconds, and then decelerates
at a rate of 2 m/s2until it comes to a stop. What is the total distance traveled
by the car during this entire motion?
Solution
Step 1: Find the distance traveled during acceleration. Given that the car
accelerates from rest at a rate of 3 m/s2for 8 seconds, we can use the kinematic
equation:
d=1
2at2
where dis the distance traveled, ais the acceleration, and tis the time. Sub-
stitute a= 3 m/s2and t= 8 s:
d=1
2×3×(8)2
d= 96 m
Step 2: Find the distance traveled at constant velocity. During the time
where the car maintains a constant velocity, the distance traveled is given by:
distance = velocity ×time
Since the car maintains a constant velocity, the distance traveled at constant
velocity is:
distance = velocity ×time = 3 m/s ×10 s = 30 m
Step 3: Find the distance traveled during deceleration. The deceleration
rate is 2 m/s2until the car comes to a stop. The distance traveled during
deceleration can be calculated using the equation:
d=v2
f−v2
i
2a
6
where vfis the final velocity, viis the initial velocity, and ais the deceleration
rate. The initial velocity during deceleration is the constant velocity of 3 m/s.
The final velocity is 0 m/s when the car comes to a stop.
d=(0)2−(3)2
2(−2)
d=−9
−4=9
4= 2.25 m
Step 4: Calculate the total distance traveled. The total distance traveled by
the car is the sum of the distances during acceleration, at constant velocity, and
during deceleration.
Total distance = 96 m + 30 m + 2.25 m = 128.25 m
Therefore, the total distance traveled by the car during this entire motion is
128.25 meters.
Question 9
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2for 10
seconds along a straight road. After this time, the car continues at a constant
speed for an additional 20 seconds. What is the total distance traveled by the
car during this time interval?
Solution
Step 1: Find the distance traveled during the acceleration phase. Given that
the car starts from rest and accelerates at a constant rate of 2.5 m/s2for 10
seconds, we can find the distance traveled during this time using the equation
for motion:
d1=1
2·a·t2
where: a= acceleration = 2.5 m/s2(constant) t= time = 10 seconds
Substitute the values into the equation:
d1=1
2·2.5·(10)2
d1= 0.5·2.5·100
d1= 125 m
Therefore, the distance traveled during the acceleration phase is 125 meters.
7
Step 2: Find the distance traveled during the constant speed phase. During
the constant speed phase, the car continues for an additional 20 seconds. Since
the speed is constant, we can find the distance traveled using the equation:
d2=v·t
where: v= constant speed
Since the car continues at a constant speed, the distance traveled during this
phase is:
d2=v·20
Step 3: Find the total distance traveled by the car. The total distance
traveled is the sum of the distances traveled during the acceleration and constant
speed phases:
Total distance = d1+d2= 125 + (v·20)
To find the constant speed v, we use the fact that the car started from rest:
v=a·t= 2.5·10 = 25 m/s
Therefore,
Total distance = 125 + (25 ·20) = 125 + 500 = 625 m
Therefore, the total distance traveled by the car during this time interval is
625 meters.
Question 10
Question
A car traveling at a constant velocity of 25 m/s passes a stationary police car.
The police car starts accelerating at a rate of 2 m/s2just as the speeding car
passes. After what time interval and at what distance will the police car catch
up to the speeding car?
Solution
Step 1: Determine the time it takes for the police car to catch up to the speeding
car. Let tbe the time it takes for the police car to catch up to the speeding car.
The position of the speeding car can be described by the equation:
xspeeding = 25t
The position of the police car can be described by the equation:
xpolice =1
2·2t2
8
Step 2: Set up an equation to find the time at which the police car catches
up to the speeding car. The police car catches up to the speeding car when the
positions of both cars are equal:
25t=1
2·2t2
Step 3: Solve for t. This equation simplifies to:
25t= 2t2
2t2−25t= 0
t(2t−25) = 0
This equation has two possible solutions: t= 0 and t= 12.5.
Step 4: Analyze the solutions. The solution t= 0 represents the initial
time when the police car starts accelerating and the speeding car passes. The
solution t= 12.5 represents the time it takes for the police car to catch up to
the speeding car.
Step 5: Find the distance at which the police car catches up to the speeding
car.
xspeeding = 25 ×12.5 = 312.5 m
Therefore, the police car catches up to the speeding car after 12.5 seconds
and at a distance of 312.5 meters from the starting point.
Question 11
Question
A car is moving along a straight road. The acceleration of the car is given by
a(t)=2t−1 m/s2, where tis the time in seconds and t≥0. If the car starts
from rest at t= 0, determine the car’s velocity and position as a function of
time.
Solution
Step 1: To find the velocity function v(t), we need to integrate the acceleration
function a(t).
a(t) = dv
dt = 2t−1
Integrating both sides with respect to t:
Zdv =Z(2t−1)dt
v(t) = Z(2t−1)dt
9
v(t) = t2−t+C1
Step 2: Since the car starts from rest, we know that v(0) = 0.
0=02−0 + C1
C1= 0
So, the velocity function is given by:
v(t) = t2−t
Step 3: To find the position function x(t), we need to integrate the velocity
function v(t).
v(t) = dx
dt =t2−t
Integrating both sides with respect to t:
Zdx =Z(t2−t)dt
x(t) = Z(t2−t)dt
x(t) = 1
3t3−1
2t2+C2
Step 4: Since the car starts from rest, we know that x(0) = 0.
0 = 1
3·03−1
2·02+C2
C2= 0
So, the position function is given by:
x(t) = 1
3t3−1
2t2
Question 12
Question
A car accelerates from rest with a constant acceleration of 4 m/s2. At the same
time, a truck 240 m ahead of the car starts moving with a constant velocity of
10 m/s. How far from the truck does the car overtake it?
10
Solution
Let’s denote the initial position of the car and the truck as xcar(0) = 0 m and
xtruck(0) = 240 m, respectively. The initial velocities are vcar(0) = 0 m/s and
vtruck(0) = 10 m/s, respectively. The acceleration of the car is acar = 4 m/s2.
The equations to find the position of the car and truck as functions of time
are:
For the car:
xcar(t) = 1
2acart2
For the truck:
xtruck(t) = vtruckt+xtruck(0)
To find the time it takes for the car to overtake the truck, we set xcar(t) =
xtruck(t):
1
2acart2=vtruckt+xtruck(0)
Solving for t, we get:
2t2−10t−240 = 0
This is a quadratic equation, whose solutions are t= 12 s (ignoring the
negative solution).
Now, let’s find how far from the truck the car overtakes it:
xovertake =xcar(t)
Step 1: Calculate the position of the car at time t= 12 s.
xcar(12) = 1
2×4×(12)2= 288 m
Therefore, the car overtakes the truck 288 m from its initial position.
Question 13
Question
A car starts from rest at a stop sign and accelerates uniformly at a rate of 2.5
m/s2in a straight line for a distance of 150 m. What is the final velocity of the
car?
Solution
Step 1: Identify the known variables. The initial velocity (v0) of the car is 0
m/s. The acceleration (a) of the car is 2.5 m/s2. The distance (d) the car travels
is 150 m. The final velocity is what we’re trying to find (vf).
11
Step 2: Choose the appropriate kinematic equation to solve for the final
velocity. Since we are not given the time taken for the car to accelerate, we can
use the kinematic equation:
v2
f=v2
0+ 2ad
Step 3: Plug in the known values into the kinematic equation and solve for
the final velocity. Substitute the values into the formula:
v2
f= (0)2+ 2(2.5)(150)
v2
f= 0 + 750
vf=√750
vf≈27.39 m/s
Step 4: State the final velocity of the car. Therefore, the final velocity of
the car is approximately 27.39 m/s.
Question 14
Question
A car starts from rest at t= 0 and accelerates uniformly at 2 m/s2for 5 sec-
onds. It then maintains a constant velocity for 10 seconds before decelerating
uniformly at 1 m/s2until it comes to a stop. What is the total distance the car
travels during this entire motion?
Solution
Step 1: Find the distance traveled during acceleration from rest at t= 0 to
t= 5 seconds. The distance traveled during acceleration can be found using the
formula: x=1
2at2, where ais the acceleration, and tis the time. Substitute
a= 2 m/s2and t= 5 seconds: x=1
2×2 m/s2×(5 s)2= 25 m.
Step 2: Find the distance traveled during constant velocity from t= 5 to
t= 15 seconds. Since the velocity is constant during this time interval, the
distance traveled is simply the product of velocity and time. The velocity after
5 seconds is 2 m/s ×5 s = 10 m.
Step 3: Find the distance traveled during deceleration from t= 15 to t= 25
seconds. The distance traveled during deceleration can be found using the
formula: x=vit+1
2at2, where viis the initial velocity, ais the acceleration,
and tis the time. The final velocity is 0 since the car comes to a stop. Substitute
vi= 10 m/s, a=−1 m/s2(negative due to deceleration), and t= 10 seconds:
x= 10 m/s ×10 s + 1
2×(−1 m/s2)×(10 s)2= 50 m.
Step 4: Calculate the total distance traveled by summing the distances from
each step. Total distance = 25 m + 10 m + 50 m = 85 m.
12
Question 15
Question
A car accelerates from rest at a constant rate of 2 m/s2for 10 seconds. After
this time, the car maintains a constant speed for 20 seconds before decelerating
to a stop at a rate of 3 m/s2. What is the total distance traveled by the car
during this entire trip?
Solution
Step 1: Find the distance traveled during acceleration. The distance traveled
during acceleration can be found using the equation:
d=1
2at2
where ais the acceleration and tis the time. Substitute a= 2 m/s2and t= 10 s:
d=1
2×2×(10)2
d=1
2×2×100
d= 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 velocity and tis the time. Since the car maintains a constant
speed during this time, vis the same as the final velocity after acceleration. The
final velocity after acceleration is:
v=a×t= 2 ×10 = 20 m/s
Substitute v= 20 m/s and t= 20 s:
d= 20 ×20
d= 400 m
Step 3: Find the distance traveled during deceleration. The distance traveled
during deceleration can also be found using the equation:
d=1
2at2
13
where ais the deceleration and tis the time. Substitute a=−3 m/s2(since it’s
deceleration) and t= 20 s:
d=1
2× −3×(20)2
d=1
2× −3×400
d=−600 m
Step 4: Calculate the total distance traveled. The total distance traveled
is the sum of the distances traveled during acceleration, constant velocity, and
deceleration:
Total distance = 100 m + 400 m −600 m
Total distance = 100 m + 400 m −600 m
Total distance = −100 m
Therefore, the total distance traveled by the car during the entire trip is 100
meters.
Question 16
Question
A car accelerates uniformly from rest at a rate of 3.00 m/s2for 8.00 seconds,
then maintains a constant velocity for 15.00 seconds, and finally decelerates
uniformly to a stop in 13.00 seconds. What is the average velocity of the car
during the entire 36.00-second trip?
Solution
Step 1: Find the distance traveled during the acceleration phase. Using the
equation for position at time tduring constant acceleration:
s=vit+1
2at2
where sis the distance traveled, viis the initial velocity, ais the acceleration,
and tis the time. Given: vi= 0 m/s, a= 3.00 m/s2,t= 8.00 s. Plugging in the
values:
s= 0 + 1
2×3.00 ×(8.00)2= 96.00 m
Step 2: Find the distance traveled during the deceleration phase. Given:
vf= 0 m/s, a=−3.00 m/s2,t= 13.00 s. Using the same formula and plugging
in the values:
s= 0 + 1
2× −3.00 ×(13.00)2=−253.50 m
14
The negative sign indicates opposite direction of motion.
Step 3: Find the average velocity during the entire trip. The total distance
covered by the car is the sum of the distances covered during acceleration,
constant velocity, and deceleration.
dtotal = 96.00 m + 0 m + (−253.50 m) = −157.50 m
The average velocity ¯vis given by:
¯v=dtotal
ttotal
where ttotal = 36.00 s. Plugging in the values:
¯v=−157.50 m
36.00 s =−4.38 m/s
Therefore, the average velocity of the car during the entire 36.00-second trip
is −4.38 m/s.
Question 17
Question
A car starts from rest and accelerates at a constant rate of 3 m/s2for a distance
of 100 meters. Calculate the final velocity of the car.
Solution
Step 1: We can use the kinematic equation relating final velocity, initial velocity,
acceleration, and displacement:
v2=u2+ 2as
where - vis the final velocity, - uis the initial velocity (in this case, the car
starts from rest so u= 0), - ais the acceleration, and - sis the displacement.
Step 2: Substituting the given values into the equation, we have:
v2= 0 + 2(3)(100)
v2= 600
Step 3: Taking the square root of both sides to find the final velocity:
v=√600
v≈24.49 m/s
Therefore, the final velocity of the car is approximately 24.49 m/s.
15
Question 18
Question
A car starts from rest and accelerates uniformly at 2.0 m/s2for 10 seconds.
Then the car moves with a constant velocity for the next 20 seconds. If the car
finally decelerates uniformly at 1.0 m/s2until it stops, find the total distance
traveled by the car during the entire motion.
Solution
Step 1: Calculate the distance traveled during acceleration phase. The dis-
tance traveled during acceleration phase can be calculated using the kinematic
equation:
d=vit+1
2at2
where: - viis the initial velocity (0 m/s), - ais the acceleration (2.0 m/s2), - t
is the time (10 seconds).
Substitute the values into the equation:
d= 0 + 1
2×2.0×(10)2= 100 m
So, the car travels 100 meters during the acceleration phase.
Step 2: Calculate the distance traveled during constant velocity phase. Since
the car moves with a constant velocity, the distance traveled is simply the prod-
uct of the velocity and time:
d=vt
where: - vis the constant velocity, - tis the time (20 seconds).
Since velocity is distance divided by time, we can rearrange the equation to
find the distance:
d=dconstant velocity
20 ×20 = dconstant velocity
Thus, the distance traveled during the constant velocity phase is dependent on
the velocity during that phase.
Step 3: Calculate the distance traveled during deceleration phase. The dis-
tance traveled during deceleration phase can be calculated using the same kine-
matic equation:
d=vit+1
2at2
where: - viis the initial velocity (the velocity after the constant velocity phase),
-ais the deceleration (-1.0 m/s2), - tis the time (unknown).
The velocity after the constant velocity phase is the same as the constant
velocity during that phase. This velocity can be calculated using the fact that
acceleration is the rate of change of velocity.
a=vf−vi
t
16
Since the car moves with constant velocity, the final velocity after 20 seconds is
the same as the constant velocity. Thus, vf=vconstant velocity. We can rearrange
the equation to find vconstant velocity in terms of acceleration:
2.0 = vconstant velocity −0
10
vconstant velocity = 20 m/s
Substitute the values into the kinematic equation:
d= 20 ×10 + 1
2×(−1.0) ×t2
20t=1
2t2
t= 40 seconds
Therefore, the distance traveled during deceleration phase is:
d= 20 ×40 + 1
2×(−1.0) ×(40)2= 800 m
Step 4: Calculate the total distance traveled. The total distance traveled is
the sum of distances traveled during each phase:
Total distance = 100 + dconstant velocity + 800
Total distance = 100 + 20 ×20 + 800 = 900 meters
Thus, the total distance traveled by the car during the entire motion is 900
meters.
Question 19
Question
A car starts from rest and accelerates along a straight road with a constant
acceleration of 3.0 m/s2. At the same instant, a truck passes the car from
behind moving with a constant velocity of 20.0 m/s.
1. How far does the car travel before it overtakes the truck?
2. How long does it take for the car to overtake the truck?
17
Solution
Let’s denote the initial position of the car and the truck as 0 meters. We will
first find the time it takes for the car to overtake the truck, and then use this
time to find the distance the car has traveled at that time.
Step 1: Find the time for the car to overtake the truck. The position
of the car as a function of time can be described by the equation:
xcar(t) = 1
2at2
Where: - a= 3.0 m/s2is the acceleration of the car. - xcar(t) is the position of
the car at time t.
The position of the truck as a function of time can be described by the
equation:
xtruck(t) = 20t
Where: - xtruck(t) is the position of the truck at time t.
The car overtakes the truck when their positions are the same:
xcar(t) = xtruck(t)
1
2·3.0t2= 20t
1.5t2= 20t
1.5t= 20
t=20
1.5= 13.33 s
Step 2: Find the distance the car travels before overtaking the
truck. Using the time found in Step 1, we can find the distance the car has
traveled at that time:
xcar(13.33) = 1
2·3.0·(13.33)2
xcar(13.33) = 89.21 m
Thus, 1. The car travels 89.21 meters before overtaking the truck. 2. It
takes 13.33 seconds for the car to overtake the truck.
Question 20
Question
A car accelerates uniformly from rest and reaches a velocity of 25 m/s in 10
seconds. Calculate the distance traveled by the car during this time.
18
Solution
Step 1: Find the acceleration of the car using the formula a=vf−vi
t, where a
is the acceleration, vfis the final velocity, viis the initial velocity, and tis the
time.
Given: vf= 25 m/s, vi= 0 m/s, t = 10 s
Acceleration: a=25 m/s −0 m/s
10 s = 2.5 m/s2
Step 2: Calculate the distance traveled by the car using the formula d=
vit+1
2at2, where dis the distance traveled, viis the initial velocity, tis the
time, and ais the acceleration.
Distance: d= 0 ×10 + 1
2×2.5×(10)2
d= 0 + 1
2×2.5×100
d=1
2×250 = 125 m
Therefore, the car travels a distance of 125 meters during the 10-second
intervals.
Question 21
Question
A car is initially at rest at t= 0 and accelerates uniformly at 2 m/s2. At t= 5
s, the velocity of the car is measured to be 10 m/s. What is the position of the
car at t= 10 s?
Solution
Step 1: Determine the acceleration of the car using the given information.
The acceleration of the car is given as 2 m/s2.
Step 2: Find the velocity of the car at t= 10 s.
Using the equation for uniformly accelerated motion:
v=u+at
where: - vis the final velocity (10 m/s), - uis the initial velocity (0 m/s), - ais
the acceleration (2 m/s2), - tis the time interval (5 s).
Substitute the known values into the formula:
10 = 0 + 2 ×5
10 = 10 m/s
19
Step 3: Calculate the displacement of the car at t= 10 s.
To find the displacement, we use the equation for displacement in uniformly
accelerated motion:
s=ut +1
2at2
where: - sis the displacement, - uis the initial velocity, - ais the acceleration,
-tis the time interval.
Substitute the values:
s= 0 ×10 + 1
2×2×102
s= 0 + 100
s= 100 m
Therefore, the position of the car at t= 10 s is 100 meters.
Question 22
Question
A car accelerates from rest at a constant rate of 4 m/s2. How far does the car
travel in the first 5 seconds?
Solution
Step 1: Determine the acceleration of the car
Given that the car accelerates at a constant rate of 4 m/s2, we have a= 4 m/s2.
Step 2: Determine the initial velocity of the car
Since the car starts from rest, the initial velocity v0is 0 m/s.
Step 3: Use the kinematic equation
The kinematic equation we can use to find the distance traveled by the car is:
x=v0t+1
2at2
Step 4: Plug in the values and solve for x
Substitute v0= 0 m/s, a= 4 m/s2, and t= 5 s into the equation:
x= 0 ×5 + 1
2×4×(5)2
x= 0 + 10 ×5
x= 50 m
Therefore, the car travels 50 meters in the first 5 seconds.
20
Question 23
Question
A car starts from rest at t= 0 and accelerates at a constant rate of 2.0 m/s2.
At what time after it started does the car’s speed reach 25 m/s?
Solution
Step 1: Let’s denote the final speed of the car as v, acceleration as a, initial
speed as u(which is 0 since the car starts from rest), and time as t. We are
given a= 2.0 m/s2and v= 25 m/s.
Step 2: We can use the kinematic equation v=u+at, where vis the final
speed, uis the initial speed, ais the acceleration, and tis the time.
Step 3: Plugging in the given values into the equation, we get:
25 = 0 + 2.0t
Step 4: Solving for t, we find:
t=25
2.0= 12.5 s
Step 5: Therefore, the car’s speed reaches 25 m/s at 12.5 seconds after it
started.
Question 24
Question
A car starts from rest and accelerates uniformly at 3.0 m/s2for 8.0 seconds.
After this time, the car continues at a constant velocity for an additional 12
seconds, before coming to a stop with a uniform acceleration of -2.0 m/s2.
Determine the total distance traveled by the car during this entire motion.
Solution
Step 1: Calculate the distance traveled during the first acceleration phase. To
find the distance traveled during the acceleration phase, we can use the equation:
d=vit+1
2at2
where viis the initial velocity, tis the time, and ais the acceleration.
Given: Initial velocity, vi= 0 m/s (car starts from rest) Acceleration, a= 3.0
m/s2Time, t= 8.0 s
Plugging in the values, we get:
d= (0 ×8.0) + 1
2×3.0×(8.0)2
21
d= 0 + 1
2×3.0×64.0
d= 0 + 96.0
d= 96.0 m
Therefore, the distance traveled during the first acceleration phase is 96.0
m.
Step 2: Calculate the distance traveled during the constant velocity phase.
During the constant velocity phase, the distance traveled is given by:
d=vt
where vis the constant velocity and tis the time.
Given: Constant velocity during this phase, v= 3.0 m/s Time, t= 12.0 s
Plugging in the values, we get:
d= 3.0×12.0
d= 36.0 m
Therefore, the distance traveled during the constant velocity phase is 36.0
m.
Step 3: Calculate the distance traveled during the deceleration phase. To
find the distance traveled during the deceleration phase, we can use the same
equation as in Step 1, but with the negative acceleration value:
d=vit+1
2at2
where viis the initial velocity (which is the constant velocity in this case), tis
the time, and ais the deceleration.
Given: Initial velocity (constant velocity), vi= 3.0 m/s Deceleration, a=
−2.0 m/s2Time, t= 8.0 s
Plugging in the values, we get:
d= 3.0×8.0 + 1
2× −2.0×(8.0)2
d= 24.0 + 1
2× −2.0×64.0
d= 24.0−64.0
d=−40.0 m
Therefore, the distance traveled during the deceleration phase is 40.0 m
(negative sign indicates the direction of travel).
Step 4: Calculate the total distance traveled by the car. The total distance
traveled by the car is the sum of the distances calculated in Step 1, Step 2, and
Step 3: Total distance = Distance in acceleration phase + Distance in constant
velocity phase + Distance in deceleration phase Total distance = 96.0 + 36.0 -
40.0 Total distance = 92.0 m
Therefore, the total distance traveled by the car during this entire motion is
92.0 meters.
22
Question 25
Question
A particle moves along the x-axis according to the equation x(t)=4t3−3t2+ 2,
where xis in meters and tis in seconds. Determine the velocity and acceleration
of the particle when t= 2 s.
Solution
Step 1: Find the velocity of the particle by taking the derivative of the position
function with respect to time.
Given x(t) = 4t3−3t2+ 2
Velocity, v(t) = dx
dt =d
dt(4t3−3t2+ 2)
v(t) = 12t2−6t
Step 2: Find the velocity of the particle when t= 2 s.
v(2) = 12(2)2−6(2)
v(2) = 48 −12
v(2) = 36 m/s
Step 3: Find the acceleration of the particle by taking the derivative of the
velocity function with respect to time.
Acceleration, a(t) = dv
dt =d
dt(12t2−6t)
a(t) = 24t−6
Step 4: Find the acceleration of the particle when 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 36 m/s and the
acceleration is 42 m/s2.
Question 26
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2along a
straight road for 10 seconds. After this time, the car continues at a constant
speed for an additional 20 seconds. Find the total distance traveled by the car
during this time interval.
23
Solution
Step 1: Find the distance traveled during the acceleration phase. Since the car
starts from rest, the initial velocity, u= 0. The acceleration, a= 3.0 m/s2. The
time for acceleration, t= 10 s.
Using the equation s=ut +1
2at2, we can find the distance traveled during
the acceleration phase.
s= (0)(10) + 1
2(3.0)(10)2
s= 0 + 1
2(3.0)(100)
s= 150 m
Step 2: Find the distance traveled during the constant speed phase. During
the constant speed phase, the car’s velocity remains constant at the final velocity
reached at the end of the acceleration phase. The final velocity, v=u+at.
v= 0 + (3.0)(10)
v= 30 m/s
The distance traveled during the constant speed phase is given by s=vt.
s= (30)(20)
s= 600 m
Step 3: Find the total distance traveled. The total distance traveled is the
sum of the distances traveled during the acceleration and constant speed phases.
Total distance = 150 m + 600 m Total distance = 750 m
Question 27
Question
A car accelerates uniformly from rest to a speed of 25 m/s over a distance of
200 m. Find the time it takes for the car to reach this speed.
Solution
Step 1: Identify the knowns and unknowns.
Given: Initial velocity, u= 0 m/s
Final velocity, v= 25 m/s
Distance traveled, s= 200 m
Unknown: Time taken to reach final speed, t
Step 2: Use the kinematic equation v=u+at where ais the acceleration.
Since the car starts from rest, u= 0, so the equation simplifies to v=at.
24
Step 3: Calculate the acceleration using the formula a=v−u
t. Substituting
the known values, we get: a=25 m/s−0 m/s
t=25 m/s
t.
Step 4: Use the kinematic equation s=ut +1
2at2. Since u= 0, the equation
simplifies to s=1
2at2.
Step 5: Substitute the values of sand ainto the equation. 200 m = 1
2×
25 m/s
t×t2=25t
2.
Step 6: Solve for tby rearranging the equation. 200 m = 25t
2
t=2×200
25 = 16 s.
Conclusion
It takes 16 seconds for the car to reach a speed of 25 m/s accelerating uniformly
over a distance of 200 m.
Question 28
Question
A race car starts from rest and accelerates uniformly to a speed of 100 m/s in
5 seconds. What is the average acceleration of the car during this time period?
Solution
Step 1: Identify the given values. The final velocity of the car, vf, is 100 m/s.
The initial velocity of the car, vi, is 0 m/s (since it starts from rest). The time
taken, t, is 5 seconds.
Step 2: Use the kinematic equation for uniform acceleration: vf=vi+at,
where ais the acceleration. Substitute the known values into the equation:
100 = 0 + a×5. Simplify the equation: 100 = 5a.
Step 3: Solve for the acceleration. Divide both sides by 5: a=100
5. There-
fore, a= 20 m/s2.
Step 4: Write the final answer. The average acceleration of the race car
during the 5-second time period is 20 m/s2.
Question 29
Question
A ball is thrown vertically upward from the ground with an initial speed of 30
m/s. Ignoring air resistance, determine the maximum height the ball reaches
above the ground.
25
Solution
Step 1: Write down the known variables and equations related to the motion
of the ball. Given: Initial velocity (u) = 30 m/s (upwards) Final velocity (v) at
maximum height = 0 m/s (when the ball momentarily stops) Acceleration (a)
=−9.81 m/s2(due to gravity, acting downwards)
The kinematic equation relating initial velocity, final velocity, acceleration,
and displacement is:
v2=u2+ 2as
Step 2: Determine the maximum height by rearranging the kinematic equa-
tion. At the maximum height, the final velocity is 0 m/s, so:
0 = (30)2+ 2(−9.81)s
0 = 900 −19.62s
19.62s= 900
s=900
19.62
Step 3: Calculate the maximum height.
s=900
19.62 ≈45.87 m
Therefore, the maximum height the ball reaches above the ground is approx-
imately 45.87 meters.
Question 30
Question
A car is initially at rest. It then accelerates uniformly along a straight road,
reaching a speed of 30 m/s in 10 seconds. What is the acceleration of the car?
Solution
Step 1: Identify the knowns
The initial velocity vi= 0 m/s, the final velocity vf= 30 m/s, and the time
t= 10 s.
Step 2: Use the kinematic equation
We can use the kinematic equation for uniformly accelerated motion:
vf=vi+at
Step 3: Substitute the known values
Substitute vi= 0 m/s, vf= 30 m/s, and t= 10 s into the equation:
30 = 0 + a×10
26
Step 4: Solve for acceleration
Solving for acceleration agives:
a=30
10 = 3 m/s2
Step 5: State the final answer
The acceleration of the car is 3 m/s2.
Question 31
Question
A car accelerates at a constant rate from rest, covering a distance of 400 m in
20 seconds. Find the speed of the car at the end of the 20 seconds.
Solution
Step 1: Identify the given quantities and the unknown. Let’s denote: - Initial
velocity of the car, vi= 0 m/s (starting from rest) - Distance covered, d= 400
m - Time taken, t= 20 s - Final velocity of the car, vf=?
Step 2: Use the kinematic equation d=vit+1
2at2to relate the given quan-
tities. The equation can be rearranged to solve for the final velocity vf:
vf=d−vit
t
Step 3: Substitute the given values into the equation.
vf=400 −0×20
20
vf=400
20
vf= 20 m/s
Answer: The speed of the car at the end of 20 seconds is 20 m/s.
Question 32
Question
A car starts from rest and accelerates uniformly to a speed of 30 m/s in 5
seconds. What is the acceleration of the car?
27
Solution
Step 1: Identify the given values and the unknown in the problem. The initial
velocity of the car, vi, is 0 m/s (starts from rest). The final velocity of the car,
vf, is 30 m/s. The time taken for the car to reach the final velocity, t, is 5
seconds. The acceleration of the car, a, is the unknown in the problem.
Step 2: Use the kinematic equation relating initial velocity, final velocity,
acceleration, and time. The kinematic equation we can use is:
vf=vi+a·t
Step 3: Substitute the known values into the kinematic equation. Substitute
vf= 30 m/s, vi= 0 m/s, and t= 5 s into the equation:
30 = 0 + a·5
Step 4: Solve for the acceleration. Simplify the equation to solve for accel-
eration:
30 = 5a
a=30
5
a= 6 m/s2
Step 5: State the final answer. The acceleration of the car is 6 m/s2.
Question 33
Question
A car travels along a straight road. Its velocity as a function of time is given
by v(t) = 6t2−4t+ 8, where vis in m/s and tis in seconds. Determine the
acceleration of the car as a function of time and find its displacement during
the time interval 1 ≤t≤3 seconds.
Solution
Step 1: Acceleration is the time derivative of velocity, so we need to find dv
dt .
Step 1: a(t) = dv
dt =d(6t2−4t+ 8)
dt = 12t−4
Step 2: To find the displacement, we need to integrate the velocity function
over the given time interval. Using the second fundamental theorem of calculus,
Step 2: Displacement = Z3
1
v(t)dt =Z3
1
(6t2−4t+ 8)dt
=2t3−2t2+ 8t3
1= (2(3)3−2(3)2+ 8(3)) −(2(1)3−2(1)2+ 8(1))
28
= (54 −18 + 24) −(2 −2 + 8) = 60 −8 = 52 meters
Therefore, the acceleration of the car as a function of time is a(t) = 12t−4
m/s2and its displacement during the time interval 1 ≤t≤3 seconds is 52
meters.
Question 34
Question
A car is traveling with a velocity given by v(t) = 10 −2tm/s, where tis the
time in seconds. Find the acceleration of the car as a function of time.
Solution
To find the acceleration of the car as a function of time, we need to differentiate
the velocity function with respect to time t.
Step 1: Determine the expression for acceleration a(t) using the given ve-
locity function.
v(t) = 10 −2t
Step 2: Differentiate the velocity function with respect to time to find the
acceleration.
a(t) = dv
dt
a(t) = d(10 −2t)
dt
a(t) = −2
Step 3: Write down the final expression for the acceleration of the car as a
function of time. So, the acceleration of the car is a constant -2 m/s2.
Question 35
Question
A car is traveling on a straight road. The car starts from rest and accelerates
uniformly at 2.0 m/s2for 10 seconds. It then maintains a constant speed for
20 seconds before decelerating uniformly at 1.5 m/s2until it comes to a stop.
What is the total distance traveled by the car during this entire process?
29
Solution
Step 1: Find the distance traveled during the acceleration phase. The formula
for distance traveled during uniform acceleration is given by:
d=1
2at2
where dis the distance traveled, ais the acceleration, and tis the time.
Given that a= 2.0m/s2and t= 10 s, we can substitute these values into
the formula to find the distance d1traveled during acceleration phase.
d1=1
2×2.0×(10)2
d1=1
2×2.0×100
d1= 100 m
Therefore, the distance traveled during the acceleration phase is 100 meters.
Step 2: Find the distance traveled during the constant speed phase. The
distance traveled during constant speed is given by:
d=vt
where dis the distance traveled, vis the constant speed, and tis the time.
Given that the car maintains a constant speed, the distance traveled during
this phase is:
d2=v×20
Since the speed is constant, the distance traveled during this phase is simply
the speed multiplied by time.
Step 3: Find the distance traveled during the deceleration phase. Similar
to the acceleration phase, the distance traveled during uniform deceleration is
given by:
d=1
2at2
Given that the car decelerates at 1.5 m/s2until it comes to a stop, we can
find the distance d3traveled during the deceleration phase.
d3=1
2×1.5×t2
Step 4: Calculate the total distance traveled by adding the distances from
each phase. The total distance traveled by the car is:
Total distance = d1+d2+d3
30
