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# Sample Kinematics Problems with Solutions

Reference > Science > Physics > Study Guide > Unit 1: Kinematics - Motion in One Direction

Following are a variety of problems involving uniformly accelerated motion along a line.  In the solution a list of known quantities will be given followed by a list of quantities wanted.  The equations to be used will be identified by number from the list below, but the algebraic work of solving the equations will be left to the student.  In most of the problems there will be a discussion of alternate methods of solution or suggestions and hints as to how to attack this particular type of problem.  Do not try to just read this section.  You must use pencil, paper and a calculator to obtain the maximum value from studying these problems.

Equation 1: vavg =
ΔX
Δt
Equation 2: aavg =
Δv
Δt
Equation 3: vavg =
vi + vf
2
Equation 4: vf = vi + aΔt
Equation 5: Δx = viΔt +
1
2
a(Δt)2
Equation 6: vf2 = vi2 + 2aΔd

## Sample Problem #1

A car starts from rest and accelerates with a uniform acceleration of 10
ft
s2
.  Calculate how long it will take for the car to reach a speed of 90
ft
s
(slightly over 60 mph), and calculate the distance the car moves during this time.

## Sample Solution #1

Given
Vi
0
a
10
ft
s2
vf
90
ft
s

Needed
ΔX
Δt

This problem can be solved by a straight forward application of equation 4 to find Δt and equation 5 to find ΔX.   The student should demonstrate that the correct answers are Δt=9 s and ΔX=405 ft.  It is a good idea to try to find the answers  using slightly different procedures whenever possible.  For example, in this problem a quick check can be made by using your calculated value for Δt, from equation 4 to find the average speed and then set X=VavgΔt to get 405 ft a different way.  A second different method would be to use equation 6 and the value of Δt obtained from equation 4 to calculate ΔX.  However, while both of the alternate techniques are good checks on the validity of your calculations it is best to avoid using an answer to one part as a basis for a second  calculation whenever possible.

## Sample Problem #2

A car moving at 30
m
s
stops with a constant acceleration in a distance of 100 m.  Calculate the acceleration and the time to stop.

## Sample Solution #2

Given
Vi
30
m
s
Vf
0
m
s
Δx
100 m

Needed
Δt
a

To solve the problem apply equation 5 to find the acceleration (a=-4.5
m
s2
).  Use the calculated value of the acceleration to find the time (Δt=6.67 s) using either equation 4 or 6 (4 is easier).

## Sample Solution #3

The acceleration is in the negative direction.  Since V was arbitrarily assigned a positive direction, the acceleration must be in the opposite direction.

## Sample Problem #4

Assume that the car in problem 2 keeps the same acceleration for an additional 5 seconds.  Find its speed and position at the end of this time.

## Sample Solution #4

Given
Vi
0
a
-4.5
m
s2
Δt
5s

Needed
Vf
X

Notice that here we are not asked for ΔX but are asked for X, the position at the end of the 5th second.  We need to be sure to specify the position unambiguously,  either with reference to the car's position at the start of problem 2 or with respect to its position at the end of problem 2 and the start of this problem.  We will place the origin of the reference system at the position of the car at the start of problem 2.  Thus at the end of problem 2 it has a position of X=100 m.    Using equation 6 with the data listed above we see that ΔX= -56.25 m.  The final position of the car is therefore 100 - 56.25 or 43.75 m.    To find the speed use equation 4 and obtain a speed of  - 22.5
m
s
.  The complete answer to the problem could be stated as follows: "The car is 43.75 m from its starting point in the direction it was originally moving. It is moving back toward the starting point at a speed of 22.5
m
s
and is accelerating toward the starting point with an acceleration of 4.5
m
s2
."  This statement is much more meaningful than simply writing X = 43.75 m and V = -22.5
m
s

## Sample Problem #5

A baseball is batted vertically with an initial speed of 45
m
s
.  Calculate:
(a)     the time before it returns to the level at which it started.
(b)     the maximum height the ball reaches.
(c)     the position, speed and acceleration 6 seconds after it is batted.
(d)     the position and speed 12 seconds after it is batted.

## Sample Solution #5

Read this problem carefully and decide the order that you will do the problem. Keep in mind the facts mentioned in the hints such as: the time to reach the top is equal to one-half the total time in the air; the speed at the top is zero; and the speed when it reaches the same level is the same as the initial speed.  In multi part problems such as this one, it will be necessary to list the given quantities for each part.  As you solve the problem, check to be sure the answers to the various sections are consistent with each other.  Before starting the problem decide which direction to call positive.  In this problem we will call the upward direction positive and the downward direction negative.

(a)

Given
Vi
45
m
s
Vf
-45
m
s
ΔX
0
a
-9.8
m
s2

Needed
Δt

This is easy.  You may use either equation 4 or equation 6.  If equation 6 is  used, you will obtain two roots, Δt=0 or 9.2 seconds.  The first root, 0,  corresponds to the first time the ball is at the starting level, when it is hit.  The second root is the desired answer, the time the ball returns to the starting level.

(b)

Given
Vi
45
m
s
Vf
0 (speed at top is zero)
a
-9.8
m
s2

Needed
Δt
ΔX

Use equation 4 to find Δt = 4.6 s.  Note that this is half the total time found in part a, so these two calculations are consistent with each other.  Use equation 5 to show that   X = 103.3 m.  As a check on your work plug the value obtained for Δt into equation 6 and see if the same value is obtained for ΔX, it should be.

(c)

Given
Vi
45
m
s
t
6 s
a
-9.8
m
s2

Needed
X
Vf

No problem here.  Use equation 4 to find that the ball has a velocity of -13.8
m
s
.  The negative sign shows that the ball is moving downward.  This is  consistent with what we found in part b, that the time to reach the top of the trajectory is  4.6 seconds.  At 6 seconds the ball will have reached the top and started moving down so its speed should be negative.Use equation 6 to find ΔX to be 93.6 m.  If we take the starting point of the motion to be at X = 0 then the position is equal to ΔX so the ball is 93.6 m above the starting point.  Again this is consistent since the displacement is less than the maximum height of 103.3 found in part b.  Finally note that equation 6 gives the change in position not the distance moved.  The ball has actually traveled further than 93.6 m.  It has traveled to the top of its flight, 103.3 m, and back down to a point 93.6 m above the ground, a distance of 9.7 m.  Thus, the total distance is 113 m.  The problem also asks for the acceleration at t = 6 seconds, but while the ball remains in free fall the acceleration is constant at -9.8
m
s2
.

(d)

Don't get caught here.  You have been asked to find the speed and position at 12 seconds.  You might be tempted to repeat the procedure of part c but if you do the answers you get will not be reasonable.  The magnitude of the velocity will be greater than 45
m
s
and the position will be negative.  This can happen only if there is a hole in the ground and the ball has fallen into it and is now below ground level.  You should immediately notice that 12 s is beyond the time that the ball takes to return to the original position.  The correct answer is that the speed and position of the ball cannot be found from the information given since the ball is no longer in free fall.

## Sample Problem #6

A ball dropped from the roof of a tall building passed a window ledge with a speed of 96
ft
s
and struck the ground 1.0 s later.
(a)     What is the height of the window?
(b)     How tall is the building?

## Sample Solution #6

(a)

Given
Vi
96
ft
s
a
32.2
ft
s2
Δt
1.0 s

Needed
ΔX

Note that since all of the motion is downward we have chosen to call the downward direction positive.  We have called the speed at the start of the last second, Vi.  ΔX can be easily found by applying equation 6.(ans: ΔX = 112 ft)  Since this is the distance the ball fell during the last second, it must be the height of the window.

(b)

There are several ways to solve this part of the problem.  One could calculate the time to fall to the window using equation 4, add 1 second to get the total time, and use the total time to find the distance using equation 6.  One could use the information known about the last second of the fall to find the final speed (use equation 5) and use this to find the total distance fallen (use equation 5 again).  Another method would be to use the speed at the start of the motion (0
m
s
), the speed at the window, and the acceleration of gravity to find the distance from the top of the building to the window.  Add 112 ft to get the total height. We must decide how we are going to solve the problem before setting up our table of known quantities.  We will do it the last way.

Given
Vi
0 (dropped)
Vf
96
ft
s
(speed at window)
a
32.2
ft
s2
(free fall)

Needed
ΔX (from top of window)

Use equation 5 to find ΔX = 144 ft.  Since this is only the height from the top of the building to the window we must add the distance to the ground, 112 ft, to get the height of the building: 256 ft.

## Questions

1.
An airplane makes a takeoff run of 810 ft and leaves the ground after 10.0 s. What was its acceleration (assumed constant) during the run? With what speed does it takeoff?
2.
A train starting from rest travels 2.40mi with uniform acceleration, attaining a final speed of 60.0mi/hr. For the next 2.00 hours the train travels at a uniform speed of 60.0mi/hr. How long does the train take to travel the first 2.40mi? What is the acceleration of the train in the first 2.40mi? What is the average speed for the entire trip?
3.
The locomotive of a freight train passes a station moving at 4.0mi/hr and accelerating at a constant rate. One minute later the caboose passes the station going 16mi/hr. What is the acceleration of the train. How long is the train?
4.
A car is going 40ft/sec and accelerates at 6.5ft/s2 for 3.0sec. What is its final speed? How far does it travel? What is the average speed during the 3.0sec?
5.
An experienced acrobat can land on the ground at a speed of about 12m/sec without injury. What is the greatest height from which he can safely jump?
6.
A man on a bridge 65m high throws a rock straight down with an initial speed of 12m/sec. Neglecting air resistance: How long will it take the rock to hit the water? How fast will it be going when it hits the water?
7.
To provide the comforts of home, a space ship accelerates in free space with the same acceleration produced by gravity on the earth. How long will it take to reach 1/10 the speed of light? How far will it have gone in that time?
8.
In an automobile accident, a car is brought to rest from 30mph (44ft/sec) in a distance of 4.0ft, mostly provided by the crumpled front end. What was the average acceleration? Compare this with the acceleration due to gravity to find the number of g's (1g = 32.2ft/s2) sustained by the safety belt the driver was, of course, wearing. How many times greater would the acceleration be if the initial velocity were 60 mph?
9.
A ball is thrown vertically upward from the top of a tower with an initial speed of 80ft/sec. What is the position of the ball relative to the top of the tower after 6.0sec?
10.
An automobile accelerates from rest at a constant rate and passes two points 60.0ft apart in 2.00sec. Its speed as it passes the second point is 50.0ft/sec. Find its acceleration and the distance of the second point from the start.
11.
An arrow is shot vertically upward with a speed of 40m/sec. How high does it rise if air resistance is neglected?
12.
With what velocity must a steel ball be thrown straight up to attain a height of 4410cm?
13.
A brick is thrown practically straight up with an initial speed of 80.0ft/sec. It lands on the edge of a roof 64.0 ft above its starting point. How high does it rise? How much time is spent on the way up? With what velocity does it land on the roof? For how long a time is it in motion?
14.
A stone is thrown vertically upward with a speed of 22.0m/sec from the top of a tower 20.0m high. When will the stone reach its maximum height? When will it pass a point 10m above the tower? When will the stone reach the ground?
15.
A rock thrown in a vertical direction from the edge of a 150m cliff struck the floor of the valley 6.00sec later. What was the initial speed of the rock? Was it up or down?
16.
A man, standing on the roof of a building 30.0m high, throws a ball vertically downward with an initial speed of 5.00m/sec. What is the speed of the ball after it has been falling for 0.500 sec? Where is the ball after 1.50 sec? What is the velocity of the ball as it strikes the ground?
17.
With what speed, in ft/sec, must a ball be thrown directly upward so that it remains in the air for 10.0 sec? What will be its speed when it hits the ground? How high does the ball rise? Assign this reference page Acceleration, Free Fall, and Problem Solving Two-Body Problems and Graphical Analysis    Like us on Facebook to get updates about new resources