PHYS 202 - GENERAL PHYSICS II -
Kinematics in One Dimension
Question Bank - Set 1
Liberty University
Question 1
Question
A car starts from rest and accelerates at a constant rate of 2 m/s2. How far
does the car travel in the first 5 seconds?
Solution
Step 1: We can use the kinematic equation d=vit+1
2at2, where dis the
distance traveled, viis the initial velocity, ais the acceleration, and tis the
time.
Step 2: Since the car starts from rest, the initial velocity vi= 0. Plugging
in the values vi= 0, a= 2 m/s2, and t= 5 s into the equation, we get:
d= 0 ×5 + 1
2×2×(5)2
Step 3: Simplifying the expression gives:
d= 0 + 1
2×2×25
d= 0 + 1 ×25 = 25 m
Step 4: Therefore, the car travels a distance of 25 meters in the first 5
seconds under the given conditions.
Question 2
Question
A car is initially at rest and accelerates uniformly to a speed of 25 m/s in 5
seconds. What is the acceleration of the car?
Solution
Step 1: Identify the given values. The initial velocity of the car, u, is 0 m/s.
The final velocity of the car, v, is 25 m/s. The time taken for the car to reach
this final velocity, t, is 5 seconds. The acceleration of the car, a, is what we
need to find.
Step 2: Use the kinematic equation relating velocity, acceleration, and time.
The kinematic equation we will use is:
v=u+at
Substitute the given values into the equation:
25 = 0 + a×5
Step 3: Solve for acceleration.
a=25
5
a= 5 m/s2
Step 4: Check the units. The unit of acceleration is in m/s
²
, which is the
correct unit for acceleration.
Therefore, the acceleration of the car is 5 m/s
²
.
Question 3
Question
A car is initially traveling at a speed of 20 m/s. It accelerates at a constant rate
of 3 m/s2for 8 seconds. What is the final velocity of the car at the end of the
8-second interval?
Solution
Step 1: Determine the acceleration of the car. Given that the car accelerates at
a constant rate of 3 m/s2, the acceleration ais 3 m/s2.
Step 2: Determine the change in velocity of the car. Using the kinematic
equation v=u+at, where: - vis the final velocity, - uis the initial velocity,
-ais the acceleration, and - tis the time interval, we can find the change in
velocity: ∆v=a·t= 3 m/s2·8 s = 24 m/s.
Step 3: Determine the final velocity of the car. The final velocity vof the car
at the end of the 8-second interval is given by: v=u+ ∆v= 20 m/s + 24 m/s =
44 m/s.
Therefore, the final velocity of the car at the end of the 8-second interval is
44 m/s.
2
Question 4
Question
A particle moves along a straight line according to the equation of motion x(t) =
2t3−3t2+ 6t+ 1, where xis in meters and tis in seconds. Determine the
displacement and distance traveled by the particle during the time interval t= 0
to t= 3 seconds.
Solution
Step 1: To find the displacement of the particle during the time interval t= 0
to t= 3 seconds, we need to evaluate x(3) −x(0).
Displacement = x(3)−x(0) = (2(3)3−3(3)2+6(3)+1)−(2(0)3−3(0)2+6(0)+1)
= (54 −27 + 18 + 1) −(0 −0 + 0 + 1) = 46 m
Step 2: To find the distance traveled by the particle during the time interval
t= 0 to t= 3 seconds, we need to calculate the total length of the curve
x(t) over this interval. We can do this by integrating the absolute value of the
velocity function.
v(t) = dx
dt = 6t2−6t+ 6
Speed = |v(t)|=|6t2−6t+ 6|
Step 3: To integrate the speed function over the interval t= 0 to t= 3
seconds, we need to break the integral into two parts where the velocity is
positive and negative, and then sum them up.
Distance = Z3
0|6t2−6t+ 6|dt
Distance = Z1
0
(6t2−6t+ 6) dt +Z3
1−(6t2−6t+ 6) dt
= 9 m
Therefore, the displacement of the particle during t= 0 to t= 3 seconds is
46 meters, and the distance traveled by the particle during that time interval is
9 meters.
Question 5
Question
A car traveling at a constant velocity of 25 m/s passes a stopped truck. At the
instant the car passes, the truck starts accelerating at a rate of 2 m/s2. How
long will it take for the truck to catch up with the car?
3
Solution
Step 1: Define the variables. Let tbe the time it takes for the truck to catch
up with the car.
Step 2: Determine the equations of motion for both the car and the truck.
The position of the car as a function of time can be described by:
xcar(t) = 25t
The position of the truck as a function of time can be described by:
xtruck(t) = 1
2(2t2)
Step 3: Set up an equation to find the time at which the truck catches up
with the car. This occurs when the positions of the car and the truck are equal:
25t= 2t2
Step 4: Solve for tby rearranging the equation:
2t2−25t= 0
2t(t−12.5) = 0
Step 5: Find the positive value of tby solving for tin the equation t−12.5 =
0. This gives t= 12.5 seconds.
Answer: It will take 12.5 seconds for the truck to catch up with the car.
Question 6
Question
A car travels along a straight road with a constant velocity of 20 m/s. At time
t= 0, the car is 50 m behind a truck that is moving along the same road in the
same direction with a constant velocity of 15 m/s. At what time will the car
overtake the truck, and how far from the starting point will this occur?
Solution
Step 1: Determine the relative velocity between the car and the truck. The
relative velocity vrel of the car with respect to the truck is the difference between
the car’s velocity and the truck’s velocity. vrel =vcar−vtruck = 20 m/s−15 m/s =
5 m/s.
Step 2: Determine the time it takes for the car to overtake the truck. Let t
be the time it takes for the car to overtake the truck. During this time, the car
will have closed the initial distance of 50 meters and some additional distance
from the truck. The additional distance closed by the car is equal to the relative
4
velocity multiplied by time: dadditional =vrel ·t= 5t. Setting this equal to the
initial distance of 50 meters, we have: 50 = 5t=⇒t= 10 s.
Step 3: Determine the distance from the starting point where the overtake
occurs. The distance dtraveled by the car when it overtakes the truck is given
by: d=vcar ·t= 20 m/s ·10 s = 200 m. Thus, the car will overtake the truck
200 meters from the starting point.
Question 7
Question
A car accelerates uniformly from rest at a rate of 3 m/s2for a distance of 100
m. What is the speed of the car after it has traveled 100 m?
Solution
Step 1: Determine the final velocity using the equation for uniformly accelerated
motion:
vf=qv2
i+ 2a·d
where - vfis the final velocity, - viis the initial velocity (0 m/s, as the car starts
from rest), - ais the acceleration (3 m/s2), - dis the distance traveled (100 m).
Step 2: Substitute the given values into the equation:
vf=p0 + 2(3)(100) = √0 + 600 = √600 ≈24.49 m/s
Therefore, the speed of the car after it has traveled 100 m is approximately
24.49 m/s.
Question 8
Question
A car traveling at a constant velocity of 25 m/s passes a street lamp. A bicyclist
traveling in the same direction has a constant velocity of 10 m/s and passes the
same lamp 5 seconds after the car passes it. How far from the lamp will the
bicyclist catch up to the car?
Solution
Step 1: Let’s denote the distance from the lamp as x. We will first find the
distance traveled by the car in 5 seconds.
Step 2: The distance traveled by the car can be calculated using the formula
d=vt, where dis the distance, vis the velocity, and tis the time.
5
Substitute v= 25 m/s and t= 5 s into the formula to find the distance
traveled by the car:
dcar = (25 m/s) ·(5 s) = 125 m
Step 3: Now, let’s find the distance between the car and the bicyclist when
the bicyclist starts after 5 seconds. This distance is given by the equation:
x= (25 m/s −10 m/s) ·5 s = 15 m/s ·5 s
x= 75 m
Step 4: At this point, the bicyclist is 75 meters behind the car. Now, both
the car and the bicyclist are moving at the same velocity of 15 m/s (25 m/s -
10 m/s). To calculate the time it takes for the bicyclist to catch up to the car,
we can use the formula:
x=vt
75 m = 15 m/s ·t
Step 5: Solve for t:
t=75 m
15 m/s = 5 s
Step 6: Finally, we can find how far from the lamp the bicyclist will catch
up to the car by using the formula:
Distance traveled by the bicyclist = 10 m/s ·5 s = 50 m
Adding the 75 meters the bicyclist was behind the car initially, the total
distance from the lamp where the bicyclist catches up to the car is:
75 m + 50 m = 125 m
Therefore, the bicyclist will catch up to the car 125 meters from the lamp.
Question 9
Question
A car accelerates from rest at a constant rate of 2.0 m/s2for a distance of 100
m. After reaching this distance, the car continues to accelerate but at a lower
rate of 1.0 m/s2. How long does it take for the car to reach a speed of 25 m/s?
6
Solution
Step 1: Determine the time it takes for the car to reach the end of the first
acceleration phase.
vf=vi+at
25 m/s = 0 + 2.0 m/s2·t
25 m/s = 2t
t=25 m/s
2 m/s2
t= 12.5 s
Step 2: Calculate the distance traveled during the first acceleration phase.
d=vit+1
2at2
d= 0 ·12.5 + 1
2·2.0 m/s2·(12.5 s)2
d= 0 + 156.25
d= 156.25 m
Step 3: Determine how far the car has left to travel after the first acceleration
phase.
∆d= 100 m −156.25 m
∆d=−56.25 m
Step 4: Determine the time it takes to reach a speed of 25 m/s after the first
acceleration phase.
vf=vi+at
25 m/s = 0 + 1.0 m/s2·t
25 m/s = t
t= 25 s
The total time for the car to reach a speed of 25 m/s is 12.5 s (first acceler-
ation phase) + 25 s (second acceleration phase) = 37.5 s.
Question 10
Question
A car initially traveling at a speed of 25 m/s accelerates uniformly at 2 m/s2
for 10 seconds. What is the final velocity of the car?
7
Solution
Let’s denote the initial velocity of the car as vi= 25 m/s, the acceleration as
a= 2 m/s2, and the time duration as t= 10 s. We need to find the final velocity
vf.
Step 1: Use the kinematic equation vf=vi+at to find the final velocity.
vf= 25 m/s + 2 m/s2×10 s
vf= 25 m/s + 20 m/s
vf= 45 m/s
Step 2: Therefore, the final velocity of the car after accelerating uniformly
at 2 m/s2for 10 seconds is 45 m/s .
Question 11
Question
A car travels along a straight road. The car’s velocity as a function of time is
given by v(t) = 8t2−2t, where vis in m/s and tis in seconds. Find the total
distance the car travels between t= 0 and t= 2 seconds.
Solution
Step 1: To find the total distance traveled by the car, we need to integrate the
absolute value of the velocity function over the interval [0,2].
Total distance = Z2
0|v(t)|dt
Step 2: First, we need to find the critical points of v(t) where v(t) = 0.
8t2−2t= 0
2t(4t−1) = 0
t= 0, t =1
4
Step 3: Split the integral depending on the sign of v(t).
Z2
0|v(t)|dt =Z1/4
0−(8t2−2t)dt +Z2
1/4
(8t2−2t)dt
Step 4: Evaluate the integral.
=−8
3t3+t21/4
0
+8
3t3−t22
1/4
8
=−8
3(1
4)3+ (1
4)2+8
3(2)3−(2)2−8
3(1
4)3−(1
4)2
=1
3+64
3−1−1
3=65
3m
Therefore, the total distance the car travels between t= 0 and t= 2 seconds
is 65
3meters.
Question 12
Question
A car accelerates along a straight road, starting from rest at time t= 0. The
acceleration of the car is given by a(t) = 2t, where ais in m/s2and tis in
seconds. Find the expression for the velocity of the car as a function of time,
v(t).
Solution
Step 1: To find the expression for the velocity of the car as a function of time,
we need to integrate the acceleration function with respect to time to get the
velocity function.
Step 2: The acceleration is given as a(t)=2t. Integrating with respect to
time gives the velocity function:
Za(t)dt =Z2t dt
v(t) = Z2t dt =t2+C
where Cis the constant of integration.
Step 3: To determine the value of the constant of integration C, we can use
the initial condition that the car starts from rest, meaning v(0) = 0. Substitut-
ing t= 0 into the velocity function:
v(0) = 02+C= 0
C= 0
Step 4: Therefore, the expression for the velocity of the car as a function of
time is:
v(t) = t2
9
Question 13
Question
An object is thrown vertically upward from the ground. Its initial velocity is
20 m/s. How long does it take for the object to reach its maximum height?
(Assume the acceleration due to gravity is −9.81 m/s2)
Solution
Step 1: We start by finding the time it takes for the object to reach its maximum
height using the kinematic equation for vertical motion:
vf=v0+at
where: - vfis the final velocity (which is 0 at the maximum height), - v0is the
initial velocity (20 m/s in this case), - ais the acceleration due to gravity (-9.81
m/s2), - tis the time.
Step 2: Substitute the known values into the equation:
0 = 20 −9.81t
Step 3: Solve for t:
9.81t= 20
t=20
9.81
t≈2.04 s
Therefore, it takes approximately 2.04 seconds for the object to reach its
maximum height.
Question 14
Question
A car initially at rest accelerates at a constant rate for 10 seconds until it reaches
a speed of 30 m/s. It then maintains this speed for 20 seconds before coming
to a stop with a constant deceleration. If the total distance traveled by the car
is 750 meters, what is the magnitude of the deceleration?
Solution
Step 1: Find the acceleration during the first 10 seconds. The final velocity of
the car, vf, is 30 m/s, the initial velocity, vi, is 0 m/s, and the time, t, is 10
seconds. We can use the equation for acceleration in one dimension:
a=vf−vi
t
10
a=30 m/s −0 m/s
10 s = 3 m/s2
Step 2: Find the distance covered during the acceleration phase. We can
use the equation:
s=vit+1
2at2
s= 0 ·10 + 1
2·3·102= 150 m
Step 3: Find the distance covered during the constant velocity phase. During
this phase, the car moves at a constant speed of 30 m/s for 20 seconds. Thus,
the distance covered is:
s=v·t= 30 ·20 = 600 m
Step 4: Find the deceleration. The total distance traveled by the car is 750
meters, the distance covered during acceleration is 150 meters, and the distance
covered during constant speed is 600 meters. The remaining distance covered
during deceleration is:
750 −150 −600 = 0
Since the car comes to a stop, the final velocity is 0. Therefore, we can use
the equation of motion for deceleration:
v2
f=v2
i+ 2as
0 = 302+ 2 ·a·s
a=−900
2s=−900
2·0=−∞m/s2
Therefore, it appears there may have been a calculation error or assumption
made in the problem statement, as the calculated deceleration is not physically
possible.
Question 15
Question
A car is initially traveling at a speed of 25 m/s. It then accelerates uniformly
at 2 m/s2for 10 seconds. What is the final velocity of the car?
Solution
Step 1: Calculate the acceleration of the car using the formula a=∆v
∆twhere a
is the acceleration, ∆vis the change in velocity, and ∆tis the change in time.
Given: ∆t= 10 s, a= 2 m/s2
11
Using a=∆v
∆t, we find: ∆v=a×∆t= 2 m/s2×10 s = 20 m/s
Step 2: Calculate the final velocity of the car using the formula vf=vi+∆v
where vfis the final velocity, viis the initial velocity, and ∆vis the change in
velocity.
Given: vi= 25 m/s, ∆v= 20 m/s
vf=vi+ ∆v= 25 m/s + 20 m/s = 45 m/s
Therefore, the final velocity of the car after accelerating uniformly at 2 m/s2
for 10 seconds is 45 m/s.
Question 16
Question
A car starts from rest and accelerates uniformly at 2.5 m/s2for 10 seconds.
After this time, the brakes are applied and the car decelerates uniformly at
3.0 m/s2until it comes to a stop. Determine the total distance the car travels
during the entire motion.
Solution
Step 1: Calculate the distance traveled during the acceleration phase.
Given acceleration, a= 2.5 m/s2
Time taken for acceleration, t1= 10 s
Initial velocity, u= 0 m/s (starting from rest)
Using the kinematic equation s=ut +1
2at2, we can find the distance traveled
during acceleration.
s1= 0 ×10 + 1
2×2.5×(10)2
s1= 0 + 125
s1= 125 m
Step 2: Calculate the distance traveled during the deceleration phase.
Given deceleration, a=−3.0 m/s2(negative because it’s deceleration)
Final velocity, v= 0 m/s (the car comes to a stop)
Again, using the kinematic equation v2=u2+ 2as, we can find the distance
traveled during deceleration.
0=02+ 2 ×(−3.0) ×s2
12
s2=0
−6.0
s2= 0 m
Step 3: Calculate the total distance traveled by summing up the distances
from the acceleration and deceleration phases.
Total distance = s1+s2
Total distance = 125 + 0
Total distance = 125 m
Therefore, the total distance the car travels during the entire motion is 125
meters.
Question 17
Question
An object is initially at rest. It undergoes acceleration for 5 seconds, then de-
celerates for the next 2 seconds until it comes to a stop. The total displacement
of the object during this time is 90 meters. If the object moves in a straight
line, what is its average velocity during the entire motion?
Solution
Step 1: Calculate the initial acceleration using the first phase of motion.
Given that the object is initially at rest, the final velocity after 5 seconds
can be calculated using the equation of motion:
vf=u+at
Where: - vf= 0 (as the object comes to rest) - u= 0 (initial velocity) - a=
acceleration
Plugging in the values and solving for a:
0 = 0 + 5a
a= 0
Therefore, the initial acceleration of the object is 0 m/s2.
Step 2: Calculate the final velocity of the object after the deceleration phase.
Given that the final velocity is 0 m/s after the 2-second deceleration, the
deceleration bcan be calculated:
vf=u+bt
0 = 0 + 2b
13
b= 0
Therefore, the deceleration is also 0 m/s2.
Step 3: Calculate the total distance traveled during the acceleration phase.
During the acceleration phase, the total distance traveled can be calculated
using the kinematic equation:
s=ut +1
2at2
Where: - s= total distance (90 meters) - u= 0 (initial velocity) - a= 0
(acceleration) - t= 5 seconds
Plugging in the values:
90 = 0 + 1
2×0×52
90 = 0
This results in a contradiction, indicating an error in our assumptions or
calculations. This suggests that the question may have been set up incorrectly,
as the given values do not lead to a consistent solution.
Therefore, the average velocity cannot be determined with the provided
information due to the inconsistency in the calculations.
Question 18
Question
A car accelerates from rest at a constant rate of 2 m/s2for 5 seconds, after
which it maintains a constant velocity for 10 seconds, and then decelerates at a
rate of 1 m/s2until it stops. Find the total distance the car travels during this
time period.
Solution
Step 1: Find the distance traveled during the acceleration phase.
The distance traveled during the acceleration phase can be calculated using
the equation:
d=1
2at2
where: a= 2 m/s2(acceleration) t= 5 s (time)
Substitute the given values into the equation:
d=1
2×2×(5)2
d=1
2×2×25
14
d= 25 m
Therefore, the distance traveled during the acceleration phase is 25 meters.
Step 2: Find the distance traveled during the constant velocity phase.
During the constant velocity phase, the distance traveled can be calculated
using the equation:
d=vt
where: v= 2 m/s (constant velocity) t= 10 s (time)
Substitute the given values into the equation:
d= 2 ×10
d= 20 m
Therefore, the distance traveled during the constant velocity phase is 20
meters.
Step 3: Find the distance traveled during the deceleration phase.
The distance traveled during the deceleration phase can be calculated using
the equation:
d=vit+1
2at2
where: vi= 2 m/s (initial velocity) a=−1 m/s2(deceleration) t= 15 s
(time)
Substitute the given values into the equation:
d= 2 ×15 + 1
2×(−1) ×(15)2
d= 30 −1
2×225
d= 30 −112.5
d=−82.5 m
Therefore, the distance traveled during the deceleration phase is -82.5 me-
ters.
Step 4: Find the total distance traveled.
To find the total distance traveled, we add up the distances traveled in each
phase:
Total distance = 25 m (acceleration) + 20 m (constant velocity) + (-82.5)
m (deceleration) Total distance = 25 m + 20 m - 82.5 m Total distance = -37.5
m
Therefore, the total distance the car travels during this time period is 37.5
meters.
15
Question 19
Question
A car is traveling along a straight road with an initial velocity of 25 m/s. The
car then accelerates at a constant rate for 10 seconds, reaching a final velocity
of 35 m/s. After that, the car decelerates at a constant rate of 2 m/s
²
until it
comes to a stop. Determine the total distance traveled by the car during this
journey.
Solution
Step 1: Calculate the distance traveled during acceleration.
vf=vi+at
Where: vf= 35 m/s (final velocity), vi= 25 m/s (initial velocity), ais the
acceleration, t= 10 s (time taken to accelerate).
Solving for the acceleration a:
35 = 25 + a(10)35 = 25 + 10a10 = 10aa = 1 m/s2
Using the formula for distance traveled during acceleration:
d=vit+1
2at2
Substitute vi= 25 m/s, a= 1 m/s
²
, and t= 10 s:
d= (25)(10) + 1
2(1)(10)2
d= 250 + 50 = 300 m
Step 2: Calculate the distance traveled during deceleration. The final veloc-
ity during deceleration is 0 m/s, and the acceleration a=−2 m/s
²
.
Using the formula for distance traveled during deceleration:
v2
f=v2
i+ 2ad
Since vf= 0, the equation simplifies to:
0 = (35)2+ 2(−2)d0 = 1225 −4d4d= 1225d= 306.25 m
Step 3: Calculate the total distance traveled by the car.
Total distance = 300 + 306.25 = 606.25 m
Therefore, the total distance traveled by the car during the journey is 606.25
meters.
16
Question 20
Question
A particle starts from rest at t= 0 and moves along the x-axis with a constant
acceleration. If the particle moves a distance of 120 m in the fourth second,
what is the acceleration of the particle?
Solution
Step 1: Let’s denote the acceleration of the particle as am/s2. Since the particle
starts from rest, its initial velocity v0= 0 m/s.
Step 2: We know that the distance traveled by a particle with constant
acceleration is given by the equation:
x(t) = v0t+1
2at2
where x(t) is the position of the particle at time t,v0is the initial velocity, ais
the acceleration, and tis the time.
Step 3: Given that the particle moves a distance of 120 m in the fourth
second, we can write the position equation as:
x(4) = 0 ·4 + 1
2a·42= 120
8a= 120
a=120
8= 15 m/s2
Step 4: Therefore, the acceleration of the particle is 15 m/s2.
Question 21
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2. How long
does it take for the car to reach a speed of 30 m/s?
Solution
Step 1: Let’s first identify the given values and the unknown in this problem.
Initial velocity (u) = 0 m/s (car starts from rest)
Acceleration (a) = 2.5 m/s2
Final velocity (v) = 30 m/s
Time taken (t) = ?
Step 2: We can use the kinematic equation relating final velocity, initial
velocity, acceleration, and time:
v=u+at
17
Substitute the known values into the equation:
30 = 0 + 2.5t
Step 3: Solve for t:
t=30
2.5= 12 s
Step 4: Therefore, it takes 12 seconds for the car to reach a speed of 30 m/s.
Question 22
Question
A particle moves along a straight line according to the equation of motion:
x(t)=3t2−2t. Determine the velocity and acceleration of the particle at time
t= 2 s.
Solution
Step 1: To find the velocity of the particle, we need to take the derivative of the
equation of motion with respect to time.
Velocity v(t) = dx
dt
v(t) = d(3t2−2t)
dt
v(t)=6t−2
Step 2: Substitute t= 2 s into the velocity equation to find the velocity at
t= 2 s.
v(2) = 6(2) −2
v(2) = 12 −2
v(2) = 10 m/s
Therefore, the velocity of the particle at t= 2 s is 10 m/s.
Step 3: To find the acceleration of the particle, we need to take the derivative
of the velocity equation with respect to time.
Acceleration a(t) = dv
dt
a(t) = d(6t−2)
dt
a(t)=6
Step 4: Substitute t= 2 s into the acceleration equation to find the acceler-
ation at t= 2 s.
a(2) = 6 m/s2
Therefore, the acceleration of the particle at t= 2 s is 6 m/s2.
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Question 23
Question
A car accelerates uniformly from rest at a rate of 3 m/s2. How long does it take
the car to reach a speed of 30 m/s?
Solution
Step 1: Let’s denote the initial velocity of the car as v0= 0 m/s, the acceleration
as a= 3 m/s2, and the final speed reached as v= 30 m/s. We want to find the
time tit takes the car to reach this final speed.
Step 2: We can use the kinematic equation that relates final velocity, initial
velocity, acceleration, and time:
v=v0+at
Step 3: Substituting the given values into the equation, we get:
30 m/s = 0 m/s + 3 m/s2·t
Step 4: Solving for t, we find:
t=30 m/s
3 m/s2= 10 s
Step 5: Therefore, it takes the car 10 seconds to reach a speed of 30 m/s.
Question 24
Question
A car accelerates from rest at a constant rate of 3 m/s2for a distance of 100
meters. After reaching a certain velocity, the car decelerates at a constant rate
of 2 m/s2until it comes to a stop. Determine the total time it takes for the car
to come to a stop.
Solution
Step 1: Let’s first calculate the time it takes for the car to reach a certain
velocity before deceleration: The final velocity vfcan be calculated using the
equation:
vf=√2a·d
where ais the acceleration and dis the distance. Plugging in a= 3 m/s2and
d= 100 m:
vf=√2·3·100 = √600 ≈24.49 m/s
19
Step 2: Next, we calculate the time taken to reach the final velocity using
the equation:
v=at
Rearranging the equation to solve for time, t:
t=v
a=24.49
3≈8.16 s
Step 3: After reaching the final velocity, the car decelerates at 2 m/s2until
it comes to a stop. The time taken to stop can be calculated using the equation:
vf=at
where vf= 0, the final velocity after deceleration. Substitute a=−2 m/s2for
deceleration:
0 = −2t
t= 0 s
Step 4: The total time taken for the car to come to a stop is the sum of the
times for acceleration and deceleration:
Total time = 8.16 s + 0 s = 8.16 s
Therefore, the total time it takes for the car to come to a stop is 8.16 seconds.
Question 25
Question
A car initially traveling at a speed of 20 m/s accelerates uniformly to a speed of
40 m/s in 5 seconds. What is the distance traveled by the car during this time?
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: vi= 20 m/s, vf= 40 m/s, t = 5 s
a=40 m/s −20 m/s
5 s =20 m/s
5 s = 4 m/s2
Step 2: Use the equation of motion d=vit+1
2at2to find the distance
traveled by the car, where dis the distance, viis the initial velocity, ais the
acceleration, and tis the time.
Given: vi= 20 m/s, a = 4 m/s2, t = 5 s
d= (20 m/s)(5 s) + 1
2(4 m/s2)(5 s)2
d= 100 m + 50 m = 150 m
Therefore, the distance traveled by the car during this time is 150 meters.
20
Question 26
Question
A car initially at rest accelerates along a straight road. It covers a distance
of 400 m in the first 20 s and a distance of 600 m in the next 30 s. Find the
acceleration of the car.
Solution
Step 1: Calculate the average velocity of the car during the first 20 s.
Average velocity = displacement
time =400 m
20 s = 20 m/s
Step 2: Calculate the acceleration of the car during the first 20 s using the
equation:
Average velocity = initial velocity + final velocity
2
20 m/s = 0 + v1
2
v1= 40 m/s
Step 3: Calculate the acceleration during the first 20 s using the equation:
a=v−u
t
a=40 m/s −0
20 s = 2 m/s2
Step 4: Calculate the average velocity of the car during the next 30 s.
Average velocity = displacement
time =600 m
30 s = 20 m/s
Step 5: Calculate the acceleration of the car during the next 30 s using the
equation:
Average velocity = initial velocity + final velocity
2
20 m/s = v1+v2
2
v1+v2= 40 m/s
Step 6: Calculate the acceleration during the next 30 s using the equation:
a=v−u
t
a=v2−40 m/s
30 s = 2 m/s2
Step 7: Since the acceleration is constant, the acceleration of the car is
2 m/s2during both time intervals.
21
Question 27
Question
A car initially traveling at 20 m/s accelerates uniformly at 2 m/s2for 10 seconds.
After this time, the brakes are applied, causing the car to decelerate at 3 m/s2.
Determine the total distance the car travels during this process.
Solution
Step 1: Let’s first find the distance traveled during the acceleration phase using
the equation:
d=vit+1
2at2
where dis the distance traveled, viis the initial velocity, ais the acceleration,
and tis the time. Plugging in the values, we get:
d= (20 m/s)(10 s) + 1
2(2 m/s2)(10 s)2
d= 200 m + (0.5)(2)(100) m
d= 200 m + 100 m = 300 m
Step 2: Next, let’s find the distance traveled during the deceleration phase.
We can use the same equation, but this time the acceleration is negative:
d=vit+1
2at2
Plugging in the values, we get:
d= (20 m/s)(10 s) + 1
2(−3 m/s2)(10 s)2
d= 200 m + (0.5)(−3)(100) m
d= 200 m −150 m = 50 m
Step 3: Finally, to find the total distance traveled, we add the distances
traveled during acceleration and deceleration:
Total distance = 300 m + 50 m = 350 m
Therefore, the car travels a total distance of 350 meters during this process.
Question 28
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 10.0
seconds. After this time, the car maintains a constant velocity for another 20.0
seconds. Finally, the car decelerates at a rate of 2.0 m/s2until it comes to a
stop. Determine the total distance traveled by the car during this entire motion.
22
Solution
Step 1: Calculate the distance traveled during the acceleration phase. The dis-
tance traveled during the acceleration phase can be calculated using the equa-
tion:
d=1
2at2
where ais the acceleration and tis the time. Substitute a= 3.0 m/s2and
t= 10.0 s into the equation:
d=1
2(3.0)(10.0)2= 150.0 m
Step 2: Calculate the distance traveled during the constant velocity phase.
The distance traveled during the constant velocity phase can be calculated using
the equation:
d=vt
where vis the constant velocity and tis the time. Since the car maintains a
constant velocity, the distance traveled is simply the product of the velocity and
time. Substitute v= 3.0×10.0 = 30 m
Step 3: Calculate the distance traveled during the deceleration phase. The
distance traveled during the deceleration phase can be calculated using the
equation:
d=vit+1
2at2
where viis the initial velocity (in this case, the constant velocity) and ais the
acceleration. Substitute vi= 30.0 m/s and a=−2.0 m/s2(negative because it’s
decelerating) and t= 10.0 into the equation:
d= 30.0×10.0 + 1
2(−2.0)(10.0)2= 300.0−100.0 = 200.0 m
Step 4: Calculate the total distance traveled by the car. The total distance
traveled by the car is the sum of the distances traveled during each phase.
Total distance = 150.0 + 30.0 + 200.0 = 380.0 m
Therefore, the total distance traveled by the car during this entire motion is
380.0 meters.
Question 29
Question
An object is launched vertically upward from the ground. The initial velocity
of the object is 12 m/s and the acceleration due to gravity is −9.81 m/s2. How
long will it take for the object to reach its maximum height?
23
Solution
Step 1: We know that at the maximum height of the object, the final velocity
will be 0 m/s since the object momentarily stops before falling back down. We
can use the kinematic equation vf=vi+at where vf= 0, vi= 12 m/s, a=
−9.81 m/s2, and we want to solve for t.
0 = 12 −9.81t
9.81t= 12
t=12
9.81
t≈1.22 s
Therefore, it will take approximately 1.22 seconds for the object to reach its
maximum height.
Question 30
Question
A particle moves along a straight line according to the equation of motion x(t) =
(6.0 m/s2)t2−(2.0 m/s3)t3, where xis in meters and tis in seconds. Determine
the velocity of the particle at t= 2.0 s.
Solution
Step 1: Find the expression for the velocity of the particle. The velocity of the
particle is given by the derivative of the position function x(t) with respect to
time:
v(t) = dx
dt
Step 2: Calculate the derivative of the position function. Differentiating
x(t)=6.0t2−2.0t3with respect to tgives:
v(t) = dx
dt = 12.0t−6.0t2
Step 3: Find the velocity at t= 2.0 s. Substitute t= 2.0 s into the expression
for velocity:
v(2.0) = 12.0(2.0) −6.0(2.0)2
v(2.0) = 24.0−24.0
v(2.0) = 0 m/s
Therefore, the velocity of the particle at t= 2.0 s is 0 m/s.
24
Question 31
Question
A stone is thrown vertically upward with an initial speed of 15 m/s. How high
does it go above its starting point? Take the acceleration due to gravity as
9.81 m/s2.
Solution
Step 1: Identify the knowns and unknowns.
Let’s denote the initial velocity of the stone as vi= 15 m/s, the acceleration
due to gravity as a=−9.81 m/s2(where the negative sign indicates that the
acceleration is directed downward), and the final velocity when the stone reaches
its highest point as vf= 0 m/s. We need to find the height the stone reaches,
denoted as h.
Step 2: Choose the appropriate kinematic equation.
In this particular scenario, the stone is thrown vertically upward and comes to
a stop at its highest point before falling back down. Therefore, we can use the
kinematic equation:
v2
f=v2
i+ 2ah
Step 3: Substitute the known values into the kinematic equation.
Substitute vi= 15 m/s, vf= 0 m/s, and a=−9.81 m/s2into the equation to
solve for h:
(0)2= (15)2+ 2(−9.81)h
Step 4: Solve for the height h.
Simplify the equation:
0 = 225 −19.62h
19.62h= 225
h=225
19.62
h≈11.47 m
Therefore, the stone reaches a height of approximately 11.47 meters above
its starting point.
Question 32
Question
A car initially traveling at 25 m/s accelerates uniformly at 3 m/s2for 10 seconds.
What is the final velocity of the car?
25
Solution
Step 1: Identify the given variables and the acceleration equation. The initial
velocity of the car, vi, is 25 m/s. The acceleration of the car, a, is 3 m/s2, and
the time interval, t, is 10 seconds. We can use the kinematic equation:
vf=vi+a·t
where vfis the final velocity.
Step 2: Substitute the known values into the equation.
vf= 25 m/s + 3 m/s2·10 s
Step 3: Calculate the final velocity of the car.
vf= 25 m/s + 30 m/s = 55 m/s
Therefore, the final velocity of the car is 55 m/s .
Question 33
Question
A car is initially at rest. It then accelerates at a constant rate of 2 m/s2for
10 seconds. After this period of acceleration, the car continues to move at a
constant velocity for 20 seconds. If the total distance travelled by the car during
this time is 300 meters, what is the total displacement of the car?
Solution
Step 1: Calculate the distance travelled during acceleration phase. The distance
travelled during acceleration phase can be calculated using the equation s=
ut +1
2at2, where sis the distance travelled, uis the initial velocity, ais the
acceleration, and tis the time. Given that the car starts from rest, u= 0.
Therefore, s= 0 + 1
2×2×(10)2.s= 0 + 1
2×2×100 = 100 meters.
Step 2: Calculate the distance travelled during constant velocity phase. Dur-
ing the constant velocity phase, the car moves at a constant speed for 20 seconds.
Since speed is constant, we can use the formula s=vt to calculate the distance
traveled. The speed vduring this phase is the final velocity after acceleration.
Using the equation of motion, v=u+at, where u= 0 and a= 2 m/s2, we have
v= 0 + 2 ×10 = 20 m/s. Therefore, s= 20 ×20 = 400 meters.
Step 3: Calculate the total displacement of the car. Displacement is a vector
quantity that accounts for the change in position from the initial to the final po-
sition. Since the car started and ended at the same point, the total displacement
is the sum of the displacements during the two phases. The displacement during
acceleration is 100 meters in the positive direction, and during constant velocity
is 400 meters in the positive direction. Therefore, the total displacement is 100
+ 400 = 500 meters.
26
Question 34
Question
A car starts from rest and accelerates in a straight line with a constant accel-
eration of 4 m/s2for 10 seconds. After this time, the car maintains a constant
velocity for another 5 seconds before coming to a stop with a constant deceler-
ation of 2 m/s2. Determine the total distance the car travels during this entire
motion.
Solution
Step 1: Find the distance traveled during acceleration.
The initial velocity vinitial is 0 m/s.
The acceleration ais 4 m/s2.
The time tis 10 s.
The distance xtraveled during acceleration can be found using the kinematic
equation:
x=vinitialt+1
2at2
x= 0 ×10 + 1
2×4×(10)2
x= 0 + 2 ×100
x= 200 m
Step 2: Find the distance traveled during constant velocity.
The velocity vis constant during this time.
The time tis 5 s.
The distance xtraveled during constant velocity is given by:
x=vt
x= 4 ×5
x= 20 m
Step 3: Find the distance traveled during deceleration.
The final velocity vfinal is 0 m/s.
The deceleration ais −2 m/s2(opposite direction to motion).
The time tis 5 s.
27
The distance xtraveled during deceleration can be found using the kinematic
equation:
x=vfinalt+1
2at2
x= 0 ×5 + 1
2×(−2) ×(5)2
x= 0 + (−1) ×25
x=−25 m
Step 4: Find the total distance traveled. The total distance traveled by the
car is the sum of the distances traveled during acceleration, constant velocity,
and deceleration.
Total distance = 200 m + 20 m + (−25) m
Total distance = 195 m
Therefore, the total distance the car travels during this entire motion is 195
meters.
Question 35
Question
A car is traveling along a straight road at a constant speed of 25 m/s. Suddenly,
the driver observes an obstruction 120 m ahead. The driver applies the brakes,
causing the car to decelerate at a rate of 4 m/s2. Will the car stop before
reaching the obstruction? If not, how far from the obstruction will the car
stop?
Solution
Step 1: Identify the given variables.
The initial velocity of the car (vi) is 25 m/s.
The acceleration of the car (a) due to braking is -4 m/s2(negative because it is
deceleration).
The distance to the obstruction (d) is 120 m.
Step 2: Determine if the car will stop before reaching the obstruction.
We will use the equation of motion:
v2
f=v2
i+ 2ad
where vfis the final velocity of the car, and since the car is stopping, vf= 0.
Substitute the known values into the equation:
0 = (25)2+ 2(−4)(d)
28
Solution
Step 1: Identify the given values. The initial velocity of the car, u, is 0 m/s.
The final velocity of the car, v, is 25 m/s. The time taken for the car to reach
this final velocity, t, is 5 seconds. The acceleration of the car, a, is what we
need to find.
Step 2: Use the kinematic equation relating velocity, acceleration, and time.
The kinematic equation we will use is:
v=u+at
Substitute the given values into the equation:
25 = 0 + a×5
Step 3: Solve for acceleration.
a=25
5
a= 5 m/s2
Step 4: Check the units. The unit of acceleration is in m/s
²
, which is the
correct unit for acceleration.
Therefore, the acceleration of the car is 5 m/s
²
.
Question 3
Question
A car is initially traveling at a speed of 20 m/s. It accelerates at a constant rate
of 3 m/s2for 8 seconds. What is the final velocity of the car at the end of the
8-second interval?
Solution
Step 1: Determine the acceleration of the car. Given that the car accelerates at
a constant rate of 3 m/s2, the acceleration ais 3 m/s2.
Step 2: Determine the change in velocity of the car. Using the kinematic
equation v=u+at, where: - vis the final velocity, - uis the initial velocity,
-ais the acceleration, and - tis the time interval, we can find the change in
velocity: ∆v=a·t= 3 m/s2·8 s = 24 m/s.
Step 3: Determine the final velocity of the car. The final velocity vof the car
at the end of the 8-second interval is given by: v=u+ ∆v= 20 m/s + 24 m/s =
44 m/s.
Therefore, the final velocity of the car at the end of the 8-second interval is
44 m/s.
2
Question 4
Question
A particle moves along a straight line according to the equation of motion x(t) =
2t3−3t2+ 6t+ 1, where xis in meters and tis in seconds. Determine the
displacement and distance traveled by the particle during the time interval t= 0
to t= 3 seconds.
Solution
Step 1: To find the displacement of the particle during the time interval t= 0
to t= 3 seconds, we need to evaluate x(3) −x(0).
Displacement = x(3)−x(0) = (2(3)3−3(3)2+6(3)+1)−(2(0)3−3(0)2+6(0)+1)
= (54 −27 + 18 + 1) −(0 −0 + 0 + 1) = 46 m
Step 2: To find the distance traveled by the particle during the time interval
t= 0 to t= 3 seconds, we need to calculate the total length of the curve
x(t) over this interval. We can do this by integrating the absolute value of the
velocity function.
v(t) = dx
dt = 6t2−6t+ 6
Speed = |v(t)|=|6t2−6t+ 6|
Step 3: To integrate the speed function over the interval t= 0 to t= 3
seconds, we need to break the integral into two parts where the velocity is
positive and negative, and then sum them up.
Distance = Z3
0|6t2−6t+ 6|dt
Distance = Z1
0
(6t2−6t+ 6) dt +Z3
1−(6t2−6t+ 6) dt
= 9 m
Therefore, the displacement of the particle during t= 0 to t= 3 seconds is
46 meters, and the distance traveled by the particle during that time interval is
9 meters.
Question 5
Question
A car traveling at a constant velocity of 25 m/s passes a stopped truck. At the
instant the car passes, the truck starts accelerating at a rate of 2 m/s2. How
long will it take for the truck to catch up with the car?
3
Solution
Step 1: Define the variables. Let tbe the time it takes for the truck to catch
up with the car.
Step 2: Determine the equations of motion for both the car and the truck.
The position of the car as a function of time can be described by:
xcar(t) = 25t
The position of the truck as a function of time can be described by:
xtruck(t) = 1
2(2t2)
Step 3: Set up an equation to find the time at which the truck catches up
with the car. This occurs when the positions of the car and the truck are equal:
25t= 2t2
Step 4: Solve for tby rearranging the equation:
2t2−25t= 0
2t(t−12.5) = 0
Step 5: Find the positive value of tby solving for tin the equation t−12.5 =
0. This gives t= 12.5 seconds.
Answer: It will take 12.5 seconds for the truck to catch up with the car.
Question 6
Question
A car travels along a straight road with a constant velocity of 20 m/s. At time
t= 0, the car is 50 m behind a truck that is moving along the same road in the
same direction with a constant velocity of 15 m/s. At what time will the car
overtake the truck, and how far from the starting point will this occur?
Solution
Step 1: Determine the relative velocity between the car and the truck. The
relative velocity vrel of the car with respect to the truck is the difference between
the car’s velocity and the truck’s velocity. vrel =vcar−vtruck = 20 m/s−15 m/s =
5 m/s.
Step 2: Determine the time it takes for the car to overtake the truck. Let t
be the time it takes for the car to overtake the truck. During this time, the car
will have closed the initial distance of 50 meters and some additional distance
from the truck. The additional distance closed by the car is equal to the relative
4
velocity multiplied by time: dadditional =vrel ·t= 5t. Setting this equal to the
initial distance of 50 meters, we have: 50 = 5t=⇒t= 10 s.
Step 3: Determine the distance from the starting point where the overtake
occurs. The distance dtraveled by the car when it overtakes the truck is given
by: d=vcar ·t= 20 m/s ·10 s = 200 m. Thus, the car will overtake the truck
200 meters from the starting point.
Question 7
Question
A car accelerates uniformly from rest at a rate of 3 m/s2for a distance of 100
m. What is the speed of the car after it has traveled 100 m?
Solution
Step 1: Determine the final velocity using the equation for uniformly accelerated
motion:
vf=qv2
i+ 2a·d
where - vfis the final velocity, - viis the initial velocity (0 m/s, as the car starts
from rest), - ais the acceleration (3 m/s2), - dis the distance traveled (100 m).
Step 2: Substitute the given values into the equation:
vf=p0 + 2(3)(100) = √0 + 600 = √600 ≈24.49 m/s
Therefore, the speed of the car after it has traveled 100 m is approximately
24.49 m/s.
Question 8
Question
A car traveling at a constant velocity of 25 m/s passes a street lamp. A bicyclist
traveling in the same direction has a constant velocity of 10 m/s and passes the
same lamp 5 seconds after the car passes it. How far from the lamp will the
bicyclist catch up to the car?
Solution
Step 1: Let’s denote the distance from the lamp as x. We will first find the
distance traveled by the car in 5 seconds.
Step 2: The distance traveled by the car can be calculated using the formula
d=vt, where dis the distance, vis the velocity, and tis the time.
5
Substitute v= 25 m/s and t= 5 s into the formula to find the distance
traveled by the car:
dcar = (25 m/s) ·(5 s) = 125 m
Step 3: Now, let’s find the distance between the car and the bicyclist when
the bicyclist starts after 5 seconds. This distance is given by the equation:
x= (25 m/s −10 m/s) ·5 s = 15 m/s ·5 s
x= 75 m
Step 4: At this point, the bicyclist is 75 meters behind the car. Now, both
the car and the bicyclist are moving at the same velocity of 15 m/s (25 m/s -
10 m/s). To calculate the time it takes for the bicyclist to catch up to the car,
we can use the formula:
x=vt
75 m = 15 m/s ·t
Step 5: Solve for t:
t=75 m
15 m/s = 5 s
Step 6: Finally, we can find how far from the lamp the bicyclist will catch
up to the car by using the formula:
Distance traveled by the bicyclist = 10 m/s ·5 s = 50 m
Adding the 75 meters the bicyclist was behind the car initially, the total
distance from the lamp where the bicyclist catches up to the car is:
75 m + 50 m = 125 m
Therefore, the bicyclist will catch up to the car 125 meters from the lamp.
Question 9
Question
A car accelerates from rest at a constant rate of 2.0 m/s2for a distance of 100
m. After reaching this distance, the car continues to accelerate but at a lower
rate of 1.0 m/s2. How long does it take for the car to reach a speed of 25 m/s?
6
Solution
Step 1: Determine the time it takes for the car to reach the end of the first
acceleration phase.
vf=vi+at
25 m/s = 0 + 2.0 m/s2·t
25 m/s = 2t
t=25 m/s
2 m/s2
t= 12.5 s
Step 2: Calculate the distance traveled during the first acceleration phase.
d=vit+1
2at2
d= 0 ·12.5 + 1
2·2.0 m/s2·(12.5 s)2
d= 0 + 156.25
d= 156.25 m
Step 3: Determine how far the car has left to travel after the first acceleration
phase.
∆d= 100 m −156.25 m
∆d=−56.25 m
Step 4: Determine the time it takes to reach a speed of 25 m/s after the first
acceleration phase.
vf=vi+at
25 m/s = 0 + 1.0 m/s2·t
25 m/s = t
t= 25 s
The total time for the car to reach a speed of 25 m/s is 12.5 s (first acceler-
ation phase) + 25 s (second acceleration phase) = 37.5 s.
Question 10
Question
A car initially traveling at a speed of 25 m/s accelerates uniformly at 2 m/s2
for 10 seconds. What is the final velocity of the car?
7
Solution
Let’s denote the initial velocity of the car as vi= 25 m/s, the acceleration as
a= 2 m/s2, and the time duration as t= 10 s. We need to find the final velocity
vf.
Step 1: Use the kinematic equation vf=vi+at to find the final velocity.
vf= 25 m/s + 2 m/s2×10 s
vf= 25 m/s + 20 m/s
vf= 45 m/s
Step 2: Therefore, the final velocity of the car after accelerating uniformly
at 2 m/s2for 10 seconds is 45 m/s .
Question 11
Question
A car travels along a straight road. The car’s velocity as a function of time is
given by v(t) = 8t2−2t, where vis in m/s and tis in seconds. Find the total
distance the car travels between t= 0 and t= 2 seconds.
Solution
Step 1: To find the total distance traveled by the car, we need to integrate the
absolute value of the velocity function over the interval [0,2].
Total distance = Z2
0|v(t)|dt
Step 2: First, we need to find the critical points of v(t) where v(t) = 0.
8t2−2t= 0
2t(4t−1) = 0
t= 0, t =1
4
Step 3: Split the integral depending on the sign of v(t).
Z2
0|v(t)|dt =Z1/4
0−(8t2−2t)dt +Z2
1/4
(8t2−2t)dt
Step 4: Evaluate the integral.
=−8
3t3+t21/4
0
+8
3t3−t22
1/4
8
=−8
3(1
4)3+ (1
4)2+8
3(2)3−(2)2−8
3(1
4)3−(1
4)2
=1
3+64
3−1−1
3=65
3m
Therefore, the total distance the car travels between t= 0 and t= 2 seconds
is 65
3meters.
Question 12
Question
A car accelerates along a straight road, starting from rest at time t= 0. The
acceleration of the car is given by a(t) = 2t, where ais in m/s2and tis in
seconds. Find the expression for the velocity of the car as a function of time,
v(t).
Solution
Step 1: To find the expression for the velocity of the car as a function of time,
we need to integrate the acceleration function with respect to time to get the
velocity function.
Step 2: The acceleration is given as a(t)=2t. Integrating with respect to
time gives the velocity function:
Za(t)dt =Z2t dt
v(t) = Z2t dt =t2+C
where Cis the constant of integration.
Step 3: To determine the value of the constant of integration C, we can use
the initial condition that the car starts from rest, meaning v(0) = 0. Substitut-
ing t= 0 into the velocity function:
v(0) = 02+C= 0
C= 0
Step 4: Therefore, the expression for the velocity of the car as a function of
time is:
v(t) = t2
9
Question 13
Question
An object is thrown vertically upward from the ground. Its initial velocity is
20 m/s. How long does it take for the object to reach its maximum height?
(Assume the acceleration due to gravity is −9.81 m/s2)
Solution
Step 1: We start by finding the time it takes for the object to reach its maximum
height using the kinematic equation for vertical motion:
vf=v0+at
where: - vfis the final velocity (which is 0 at the maximum height), - v0is the
initial velocity (20 m/s in this case), - ais the acceleration due to gravity (-9.81
m/s2), - tis the time.
Step 2: Substitute the known values into the equation:
0 = 20 −9.81t
Step 3: Solve for t:
9.81t= 20
t=20
9.81
t≈2.04 s
Therefore, it takes approximately 2.04 seconds for the object to reach its
maximum height.
Question 14
Question
A car initially at rest accelerates at a constant rate for 10 seconds until it reaches
a speed of 30 m/s. It then maintains this speed for 20 seconds before coming
to a stop with a constant deceleration. If the total distance traveled by the car
is 750 meters, what is the magnitude of the deceleration?
Solution
Step 1: Find the acceleration during the first 10 seconds. The final velocity of
the car, vf, is 30 m/s, the initial velocity, vi, is 0 m/s, and the time, t, is 10
seconds. We can use the equation for acceleration in one dimension:
a=vf−vi
t
10
a=30 m/s −0 m/s
10 s = 3 m/s2
Step 2: Find the distance covered during the acceleration phase. We can
use the equation:
s=vit+1
2at2
s= 0 ·10 + 1
2·3·102= 150 m
Step 3: Find the distance covered during the constant velocity phase. During
this phase, the car moves at a constant speed of 30 m/s for 20 seconds. Thus,
the distance covered is:
s=v·t= 30 ·20 = 600 m
Step 4: Find the deceleration. The total distance traveled by the car is 750
meters, the distance covered during acceleration is 150 meters, and the distance
covered during constant speed is 600 meters. The remaining distance covered
during deceleration is:
750 −150 −600 = 0
Since the car comes to a stop, the final velocity is 0. Therefore, we can use
the equation of motion for deceleration:
v2
f=v2
i+ 2as
0 = 302+ 2 ·a·s
a=−900
2s=−900
2·0=−∞m/s2
Therefore, it appears there may have been a calculation error or assumption
made in the problem statement, as the calculated deceleration is not physically
possible.
Question 15
Question
A car is initially traveling at a speed of 25 m/s. It then accelerates uniformly
at 2 m/s2for 10 seconds. What is the final velocity of the car?
Solution
Step 1: Calculate the acceleration of the car using the formula a=∆v
∆twhere a
is the acceleration, ∆vis the change in velocity, and ∆tis the change in time.
Given: ∆t= 10 s, a= 2 m/s2
11
Using a=∆v
∆t, we find: ∆v=a×∆t= 2 m/s2×10 s = 20 m/s
Step 2: Calculate the final velocity of the car using the formula vf=vi+∆v
where vfis the final velocity, viis the initial velocity, and ∆vis the change in
velocity.
Given: vi= 25 m/s, ∆v= 20 m/s
vf=vi+ ∆v= 25 m/s + 20 m/s = 45 m/s
Therefore, the final velocity of the car after accelerating uniformly at 2 m/s2
for 10 seconds is 45 m/s.
Question 16
Question
A car starts from rest and accelerates uniformly at 2.5 m/s2for 10 seconds.
After this time, the brakes are applied and the car decelerates uniformly at
3.0 m/s2until it comes to a stop. Determine the total distance the car travels
during the entire motion.
Solution
Step 1: Calculate the distance traveled during the acceleration phase.
Given acceleration, a= 2.5 m/s2
Time taken for acceleration, t1= 10 s
Initial velocity, u= 0 m/s (starting from rest)
Using the kinematic equation s=ut +1
2at2, we can find the distance traveled
during acceleration.
s1= 0 ×10 + 1
2×2.5×(10)2
s1= 0 + 125
s1= 125 m
Step 2: Calculate the distance traveled during the deceleration phase.
Given deceleration, a=−3.0 m/s2(negative because it’s deceleration)
Final velocity, v= 0 m/s (the car comes to a stop)
Again, using the kinematic equation v2=u2+ 2as, we can find the distance
traveled during deceleration.
0=02+ 2 ×(−3.0) ×s2
12
s2=0
−6.0
s2= 0 m
Step 3: Calculate the total distance traveled by summing up the distances
from the acceleration and deceleration phases.
Total distance = s1+s2
Total distance = 125 + 0
Total distance = 125 m
Therefore, the total distance the car travels during the entire motion is 125
meters.
Question 17
Question
An object is initially at rest. It undergoes acceleration for 5 seconds, then de-
celerates for the next 2 seconds until it comes to a stop. The total displacement
of the object during this time is 90 meters. If the object moves in a straight
line, what is its average velocity during the entire motion?
Solution
Step 1: Calculate the initial acceleration using the first phase of motion.
Given that the object is initially at rest, the final velocity after 5 seconds
can be calculated using the equation of motion:
vf=u+at
Where: - vf= 0 (as the object comes to rest) - u= 0 (initial velocity) - a=
acceleration
Plugging in the values and solving for a:
0 = 0 + 5a
a= 0
Therefore, the initial acceleration of the object is 0 m/s2.
Step 2: Calculate the final velocity of the object after the deceleration phase.
Given that the final velocity is 0 m/s after the 2-second deceleration, the
deceleration bcan be calculated:
vf=u+bt
0 = 0 + 2b
13
b= 0
Therefore, the deceleration is also 0 m/s2.
Step 3: Calculate the total distance traveled during the acceleration phase.
During the acceleration phase, the total distance traveled can be calculated
using the kinematic equation:
s=ut +1
2at2
Where: - s= total distance (90 meters) - u= 0 (initial velocity) - a= 0
(acceleration) - t= 5 seconds
Plugging in the values:
90 = 0 + 1
2×0×52
90 = 0
This results in a contradiction, indicating an error in our assumptions or
calculations. This suggests that the question may have been set up incorrectly,
as the given values do not lead to a consistent solution.
Therefore, the average velocity cannot be determined with the provided
information due to the inconsistency in the calculations.
Question 18
Question
A car accelerates from rest at a constant rate of 2 m/s2for 5 seconds, after
which it maintains a constant velocity for 10 seconds, and then decelerates at a
rate of 1 m/s2until it stops. Find the total distance the car travels during this
time period.
Solution
Step 1: Find the distance traveled during the acceleration phase.
The distance traveled during the acceleration phase can be calculated using
the equation:
d=1
2at2
where: a= 2 m/s2(acceleration) t= 5 s (time)
Substitute the given values into the equation:
d=1
2×2×(5)2
d=1
2×2×25
14
d= 25 m
Therefore, the distance traveled during the acceleration phase is 25 meters.
Step 2: Find the distance traveled during the constant velocity phase.
During the constant velocity phase, the distance traveled can be calculated
using the equation:
d=vt
where: v= 2 m/s (constant velocity) t= 10 s (time)
Substitute the given values into the equation:
d= 2 ×10
d= 20 m
Therefore, the distance traveled during the constant velocity phase is 20
meters.
Step 3: Find the distance traveled during the deceleration phase.
The distance traveled during the deceleration phase can be calculated using
the equation:
d=vit+1
2at2
where: vi= 2 m/s (initial velocity) a=−1 m/s2(deceleration) t= 15 s
(time)
Substitute the given values into the equation:
d= 2 ×15 + 1
2×(−1) ×(15)2
d= 30 −1
2×225
d= 30 −112.5
d=−82.5 m
Therefore, the distance traveled during the deceleration phase is -82.5 me-
ters.
Step 4: Find the total distance traveled.
To find the total distance traveled, we add up the distances traveled in each
phase:
Total distance = 25 m (acceleration) + 20 m (constant velocity) + (-82.5)
m (deceleration) Total distance = 25 m + 20 m - 82.5 m Total distance = -37.5
m
Therefore, the total distance the car travels during this time period is 37.5
meters.
15
Question 19
Question
A car is traveling along a straight road with an initial velocity of 25 m/s. The
car then accelerates at a constant rate for 10 seconds, reaching a final velocity
of 35 m/s. After that, the car decelerates at a constant rate of 2 m/s
²
until it
comes to a stop. Determine the total distance traveled by the car during this
journey.
Solution
Step 1: Calculate the distance traveled during acceleration.
vf=vi+at
Where: vf= 35 m/s (final velocity), vi= 25 m/s (initial velocity), ais the
acceleration, t= 10 s (time taken to accelerate).
Solving for the acceleration a:
35 = 25 + a(10)35 = 25 + 10a10 = 10aa = 1 m/s2
Using the formula for distance traveled during acceleration:
d=vit+1
2at2
Substitute vi= 25 m/s, a= 1 m/s
²
, and t= 10 s:
d= (25)(10) + 1
2(1)(10)2
d= 250 + 50 = 300 m
Step 2: Calculate the distance traveled during deceleration. The final veloc-
ity during deceleration is 0 m/s, and the acceleration a=−2 m/s
²
.
Using the formula for distance traveled during deceleration:
v2
f=v2
i+ 2ad
Since vf= 0, the equation simplifies to:
0 = (35)2+ 2(−2)d0 = 1225 −4d4d= 1225d= 306.25 m
Step 3: Calculate the total distance traveled by the car.
Total distance = 300 + 306.25 = 606.25 m
Therefore, the total distance traveled by the car during the journey is 606.25
meters.
16
Question 20
Question
A particle starts from rest at t= 0 and moves along the x-axis with a constant
acceleration. If the particle moves a distance of 120 m in the fourth second,
what is the acceleration of the particle?
Solution
Step 1: Let’s denote the acceleration of the particle as am/s2. Since the particle
starts from rest, its initial velocity v0= 0 m/s.
Step 2: We know that the distance traveled by a particle with constant
acceleration is given by the equation:
x(t) = v0t+1
2at2
where x(t) is the position of the particle at time t,v0is the initial velocity, ais
the acceleration, and tis the time.
Step 3: Given that the particle moves a distance of 120 m in the fourth
second, we can write the position equation as:
x(4) = 0 ·4 + 1
2a·42= 120
8a= 120
a=120
8= 15 m/s2
Step 4: Therefore, the acceleration of the particle is 15 m/s2.
Question 21
Question
A car starts from rest and accelerates at a constant rate of 2.5 m/s2. How long
does it take for the car to reach a speed of 30 m/s?
Solution
Step 1: Let’s first identify the given values and the unknown in this problem.
Initial velocity (u) = 0 m/s (car starts from rest)
Acceleration (a) = 2.5 m/s2
Final velocity (v) = 30 m/s
Time taken (t) = ?
Step 2: We can use the kinematic equation relating final velocity, initial
velocity, acceleration, and time:
v=u+at
17
Substitute the known values into the equation:
30 = 0 + 2.5t
Step 3: Solve for t:
t=30
2.5= 12 s
Step 4: Therefore, it takes 12 seconds for the car to reach a speed of 30 m/s.
Question 22
Question
A particle moves along a straight line according to the equation of motion:
x(t)=3t2−2t. Determine the velocity and acceleration of the particle at time
t= 2 s.
Solution
Step 1: To find the velocity of the particle, we need to take the derivative of the
equation of motion with respect to time.
Velocity v(t) = dx
dt
v(t) = d(3t2−2t)
dt
v(t)=6t−2
Step 2: Substitute t= 2 s into the velocity equation to find the velocity at
t= 2 s.
v(2) = 6(2) −2
v(2) = 12 −2
v(2) = 10 m/s
Therefore, the velocity of the particle at t= 2 s is 10 m/s.
Step 3: To find the acceleration of the particle, we need to take the derivative
of the velocity equation with respect to time.
Acceleration a(t) = dv
dt
a(t) = d(6t−2)
dt
a(t)=6
Step 4: Substitute t= 2 s into the acceleration equation to find the acceler-
ation at t= 2 s.
a(2) = 6 m/s2
Therefore, the acceleration of the particle at t= 2 s is 6 m/s2.
18
Question 23
Question
A car accelerates uniformly from rest at a rate of 3 m/s2. How long does it take
the car to reach a speed of 30 m/s?
Solution
Step 1: Let’s denote the initial velocity of the car as v0= 0 m/s, the acceleration
as a= 3 m/s2, and the final speed reached as v= 30 m/s. We want to find the
time tit takes the car to reach this final speed.
Step 2: We can use the kinematic equation that relates final velocity, initial
velocity, acceleration, and time:
v=v0+at
Step 3: Substituting the given values into the equation, we get:
30 m/s = 0 m/s + 3 m/s2·t
Step 4: Solving for t, we find:
t=30 m/s
3 m/s2= 10 s
Step 5: Therefore, it takes the car 10 seconds to reach a speed of 30 m/s.
Question 24
Question
A car accelerates from rest at a constant rate of 3 m/s2for a distance of 100
meters. After reaching a certain velocity, the car decelerates at a constant rate
of 2 m/s2until it comes to a stop. Determine the total time it takes for the car
to come to a stop.
Solution
Step 1: Let’s first calculate the time it takes for the car to reach a certain
velocity before deceleration: The final velocity vfcan be calculated using the
equation:
vf=√2a·d
where ais the acceleration and dis the distance. Plugging in a= 3 m/s2and
d= 100 m:
vf=√2·3·100 = √600 ≈24.49 m/s
19
Step 2: Next, we calculate the time taken to reach the final velocity using
the equation:
v=at
Rearranging the equation to solve for time, t:
t=v
a=24.49
3≈8.16 s
Step 3: After reaching the final velocity, the car decelerates at 2 m/s2until
it comes to a stop. The time taken to stop can be calculated using the equation:
vf=at
where vf= 0, the final velocity after deceleration. Substitute a=−2 m/s2for
deceleration:
0 = −2t
t= 0 s
Step 4: The total time taken for the car to come to a stop is the sum of the
times for acceleration and deceleration:
Total time = 8.16 s + 0 s = 8.16 s
Therefore, the total time it takes for the car to come to a stop is 8.16 seconds.
Question 25
Question
A car initially traveling at a speed of 20 m/s accelerates uniformly to a speed of
40 m/s in 5 seconds. What is the distance traveled by the car during this time?
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: vi= 20 m/s, vf= 40 m/s, t = 5 s
a=40 m/s −20 m/s
5 s =20 m/s
5 s = 4 m/s2
Step 2: Use the equation of motion d=vit+1
2at2to find the distance
traveled by the car, where dis the distance, viis the initial velocity, ais the
acceleration, and tis the time.
Given: vi= 20 m/s, a = 4 m/s2, t = 5 s
d= (20 m/s)(5 s) + 1
2(4 m/s2)(5 s)2
d= 100 m + 50 m = 150 m
Therefore, the distance traveled by the car during this time is 150 meters.
20
Question 26
Question
A car initially at rest accelerates along a straight road. It covers a distance
of 400 m in the first 20 s and a distance of 600 m in the next 30 s. Find the
acceleration of the car.
Solution
Step 1: Calculate the average velocity of the car during the first 20 s.
Average velocity = displacement
time =400 m
20 s = 20 m/s
Step 2: Calculate the acceleration of the car during the first 20 s using the
equation:
Average velocity = initial velocity + final velocity
2
20 m/s = 0 + v1
2
v1= 40 m/s
Step 3: Calculate the acceleration during the first 20 s using the equation:
a=v−u
t
a=40 m/s −0
20 s = 2 m/s2
Step 4: Calculate the average velocity of the car during the next 30 s.
Average velocity = displacement
time =600 m
30 s = 20 m/s
Step 5: Calculate the acceleration of the car during the next 30 s using the
equation:
Average velocity = initial velocity + final velocity
2
20 m/s = v1+v2
2
v1+v2= 40 m/s
Step 6: Calculate the acceleration during the next 30 s using the equation:
a=v−u
t
a=v2−40 m/s
30 s = 2 m/s2
Step 7: Since the acceleration is constant, the acceleration of the car is
2 m/s2during both time intervals.
21
Question 27
Question
A car initially traveling at 20 m/s accelerates uniformly at 2 m/s2for 10 seconds.
After this time, the brakes are applied, causing the car to decelerate at 3 m/s2.
Determine the total distance the car travels during this process.
Solution
Step 1: Let’s first find the distance traveled during the acceleration phase using
the equation:
d=vit+1
2at2
where dis the distance traveled, viis the initial velocity, ais the acceleration,
and tis the time. Plugging in the values, we get:
d= (20 m/s)(10 s) + 1
2(2 m/s2)(10 s)2
d= 200 m + (0.5)(2)(100) m
d= 200 m + 100 m = 300 m
Step 2: Next, let’s find the distance traveled during the deceleration phase.
We can use the same equation, but this time the acceleration is negative:
d=vit+1
2at2
Plugging in the values, we get:
d= (20 m/s)(10 s) + 1
2(−3 m/s2)(10 s)2
d= 200 m + (0.5)(−3)(100) m
d= 200 m −150 m = 50 m
Step 3: Finally, to find the total distance traveled, we add the distances
traveled during acceleration and deceleration:
Total distance = 300 m + 50 m = 350 m
Therefore, the car travels a total distance of 350 meters during this process.
Question 28
Question
A car starts from rest and accelerates at a constant rate of 3.0 m/s2for 10.0
seconds. After this time, the car maintains a constant velocity for another 20.0
seconds. Finally, the car decelerates at a rate of 2.0 m/s2until it comes to a
stop. Determine the total distance traveled by the car during this entire motion.
22
Solution
Step 1: Calculate the distance traveled during the acceleration phase. The dis-
tance traveled during the acceleration phase can be calculated using the equa-
tion:
d=1
2at2
where ais the acceleration and tis the time. Substitute a= 3.0 m/s2and
t= 10.0 s into the equation:
d=1
2(3.0)(10.0)2= 150.0 m
Step 2: Calculate the distance traveled during the constant velocity phase.
The distance traveled during the constant velocity phase can be calculated using
the equation:
d=vt
where vis the constant velocity and tis the time. Since the car maintains a
constant velocity, the distance traveled is simply the product of the velocity and
time. Substitute v= 3.0×10.0 = 30 m
Step 3: Calculate the distance traveled during the deceleration phase. The
distance traveled during the deceleration phase can be calculated using the
equation:
d=vit+1
2at2
where viis the initial velocity (in this case, the constant velocity) and ais the
acceleration. Substitute vi= 30.0 m/s and a=−2.0 m/s2(negative because it’s
decelerating) and t= 10.0 into the equation:
d= 30.0×10.0 + 1
2(−2.0)(10.0)2= 300.0−100.0 = 200.0 m
Step 4: Calculate the total distance traveled by the car. The total distance
traveled by the car is the sum of the distances traveled during each phase.
Total distance = 150.0 + 30.0 + 200.0 = 380.0 m
Therefore, the total distance traveled by the car during this entire motion is
380.0 meters.
Question 29
Question
An object is launched vertically upward from the ground. The initial velocity
of the object is 12 m/s and the acceleration due to gravity is −9.81 m/s2. How
long will it take for the object to reach its maximum height?
23
Solution
Step 1: We know that at the maximum height of the object, the final velocity
will be 0 m/s since the object momentarily stops before falling back down. We
can use the kinematic equation vf=vi+at where vf= 0, vi= 12 m/s, a=
−9.81 m/s2, and we want to solve for t.
0 = 12 −9.81t
9.81t= 12
t=12
9.81
t≈1.22 s
Therefore, it will take approximately 1.22 seconds for the object to reach its
maximum height.
Question 30
Question
A particle moves along a straight line according to the equation of motion x(t) =
(6.0 m/s2)t2−(2.0 m/s3)t3, where xis in meters and tis in seconds. Determine
the velocity of the particle at t= 2.0 s.
Solution
Step 1: Find the expression for the velocity of the particle. The velocity of the
particle is given by the derivative of the position function x(t) with respect to
time:
v(t) = dx
dt
Step 2: Calculate the derivative of the position function. Differentiating
x(t)=6.0t2−2.0t3with respect to tgives:
v(t) = dx
dt = 12.0t−6.0t2
Step 3: Find the velocity at t= 2.0 s. Substitute t= 2.0 s into the expression
for velocity:
v(2.0) = 12.0(2.0) −6.0(2.0)2
v(2.0) = 24.0−24.0
v(2.0) = 0 m/s
Therefore, the velocity of the particle at t= 2.0 s is 0 m/s.
24
Question 31
Question
A stone is thrown vertically upward with an initial speed of 15 m/s. How high
does it go above its starting point? Take the acceleration due to gravity as
9.81 m/s2.
Solution
Step 1: Identify the knowns and unknowns.
Let’s denote the initial velocity of the stone as vi= 15 m/s, the acceleration
due to gravity as a=−9.81 m/s2(where the negative sign indicates that the
acceleration is directed downward), and the final velocity when the stone reaches
its highest point as vf= 0 m/s. We need to find the height the stone reaches,
denoted as h.
Step 2: Choose the appropriate kinematic equation.
In this particular scenario, the stone is thrown vertically upward and comes to
a stop at its highest point before falling back down. Therefore, we can use the
kinematic equation:
v2
f=v2
i+ 2ah
Step 3: Substitute the known values into the kinematic equation.
Substitute vi= 15 m/s, vf= 0 m/s, and a=−9.81 m/s2into the equation to
solve for h:
(0)2= (15)2+ 2(−9.81)h
Step 4: Solve for the height h.
Simplify the equation:
0 = 225 −19.62h
19.62h= 225
h=225
19.62
h≈11.47 m
Therefore, the stone reaches a height of approximately 11.47 meters above
its starting point.
Question 32
Question
A car initially traveling at 25 m/s accelerates uniformly at 3 m/s2for 10 seconds.
What is the final velocity of the car?
25
Solution
Step 1: Identify the given variables and the acceleration equation. The initial
velocity of the car, vi, is 25 m/s. The acceleration of the car, a, is 3 m/s2, and
the time interval, t, is 10 seconds. We can use the kinematic equation:
vf=vi+a·t
where vfis the final velocity.
Step 2: Substitute the known values into the equation.
vf= 25 m/s + 3 m/s2·10 s
Step 3: Calculate the final velocity of the car.
vf= 25 m/s + 30 m/s = 55 m/s
Therefore, the final velocity of the car is 55 m/s .
Question 33
Question
A car is initially at rest. It then accelerates at a constant rate of 2 m/s2for
10 seconds. After this period of acceleration, the car continues to move at a
constant velocity for 20 seconds. If the total distance travelled by the car during
this time is 300 meters, what is the total displacement of the car?
Solution
Step 1: Calculate the distance travelled during acceleration phase. The distance
travelled during acceleration phase can be calculated using the equation s=
ut +1
2at2, where sis the distance travelled, uis the initial velocity, ais the
acceleration, and tis the time. Given that the car starts from rest, u= 0.
Therefore, s= 0 + 1
2×2×(10)2.s= 0 + 1
2×2×100 = 100 meters.
Step 2: Calculate the distance travelled during constant velocity phase. Dur-
ing the constant velocity phase, the car moves at a constant speed for 20 seconds.
Since speed is constant, we can use the formula s=vt to calculate the distance
traveled. The speed vduring this phase is the final velocity after acceleration.
Using the equation of motion, v=u+at, where u= 0 and a= 2 m/s2, we have
v= 0 + 2 ×10 = 20 m/s. Therefore, s= 20 ×20 = 400 meters.
Step 3: Calculate the total displacement of the car. Displacement is a vector
quantity that accounts for the change in position from the initial to the final po-
sition. Since the car started and ended at the same point, the total displacement
is the sum of the displacements during the two phases. The displacement during
acceleration is 100 meters in the positive direction, and during constant velocity
is 400 meters in the positive direction. Therefore, the total displacement is 100
+ 400 = 500 meters.
26
Question 34
Question
A car starts from rest and accelerates in a straight line with a constant accel-
eration of 4 m/s2for 10 seconds. After this time, the car maintains a constant
velocity for another 5 seconds before coming to a stop with a constant deceler-
ation of 2 m/s2. Determine the total distance the car travels during this entire
motion.
Solution
Step 1: Find the distance traveled during acceleration.
The initial velocity vinitial is 0 m/s.
The acceleration ais 4 m/s2.
The time tis 10 s.
The distance xtraveled during acceleration can be found using the kinematic
equation:
x=vinitialt+1
2at2
x= 0 ×10 + 1
2×4×(10)2
x= 0 + 2 ×100
x= 200 m
Step 2: Find the distance traveled during constant velocity.
The velocity vis constant during this time.
The time tis 5 s.
The distance xtraveled during constant velocity is given by:
x=vt
x= 4 ×5
x= 20 m
Step 3: Find the distance traveled during deceleration.
The final velocity vfinal is 0 m/s.
The deceleration ais −2 m/s2(opposite direction to motion).
The time tis 5 s.
27
The distance xtraveled during deceleration can be found using the kinematic
equation:
x=vfinalt+1
2at2
x= 0 ×5 + 1
2×(−2) ×(5)2
x= 0 + (−1) ×25
x=−25 m
Step 4: Find the total distance traveled. The total distance traveled by the
car is the sum of the distances traveled during acceleration, constant velocity,
and deceleration.
Total distance = 200 m + 20 m + (−25) m
Total distance = 195 m
Therefore, the total distance the car travels during this entire motion is 195
meters.
Question 35
Question
A car is traveling along a straight road at a constant speed of 25 m/s. Suddenly,
the driver observes an obstruction 120 m ahead. The driver applies the brakes,
causing the car to decelerate at a rate of 4 m/s2. Will the car stop before
reaching the obstruction? If not, how far from the obstruction will the car
stop?
Solution
Step 1: Identify the given variables.
The initial velocity of the car (vi) is 25 m/s.
The acceleration of the car (a) due to braking is -4 m/s2(negative because it is
deceleration).
The distance to the obstruction (d) is 120 m.
Step 2: Determine if the car will stop before reaching the obstruction.
We will use the equation of motion:
v2
f=v2
i+ 2ad
where vfis the final velocity of the car, and since the car is stopping, vf= 0.
Substitute the known values into the equation:
0 = (25)2+ 2(−4)(d)
28
0 = 625 −8d
8d= 625
d=625
8
d≈78.125 m
Step 3: Conclusion.
The car will not stop before reaching the obstruction. It will stop approximately
78.125 meters from the obstruction.
29
