Chapter 2 : 1-D Kinematics: Velocity & Acceleration Hints & Answers. –Updated 9/6/11
1. a) Position is the distance from a predetermined origin; displacement is the distance from the initial to
the final position. The origin is arbitrary and chosen for convenience, so position vectors can vary, but
displacement vectors don’t depend on the particular origin chosen in a problem.
b) Instantaneous velocity is the rate of motion at one particular instant of time (as calculated from the
limit of x/t =dx/dt). Average velocity is the overall displacement traveled over the total time interval that
it took. Both are the same when the velocity is constant.
c) instant velocity, change in velocity and acceleration …refer to the definitions in text and/or notes
d) constant velocity, constant & non-const acceleration …refer to the definitions in text and/or notes
e) +a is acceleration in the positive direction, -a is acceleration in the negative direction. Referring to
speeds (regardless of direction), “acceleration” means the speed is increasing, while “deceleration”
means the speed is decreasing. You should understand that a +a can accelerate or decelerate an object,
and that a –a can also accelerate or decelerate an object. The key point is that objects “speed up” when
the velocity and acceleration vectors are in the same direction (both +, or both-), and that objects “slow
down” when the velocity and acceleration vectors are in opposite directions (+/- , or -/+)
2. In preparation for the “Graphing Constant Acceleration” lab, sketch rough graphs of position, velocity,
and acceleration vs. time for the following cases of constant acceleration:
a) a=0, v>0; and also a=0, v<0; b) a>0, vo>0; and also a>0, vo<0;
c) a<0, vo>0; and also a<0, vo<0
Answer these questions about the graphs you sketched:
d) In which of the cases above was the speed (magnitude of velocity) increasing?...speed decreasing?
e) In which of the cases above is it possible for the object turn around?
f) Is it possible for the speed to be zero and the acceleration to be non-zero at any time? Explain.
You will obtain these graphs during the lab “Graphing Constant Acceleration”. So come back to this
problem after the lab. One of the points of the lab is to realize that speed increases (“acceleration”) when
velocity and acceleration are in the same direction (+/+or -/-), whereas speed decreases (“deceleration”)
when velocity and acceleration are in opposite directions (+/-or -/+).
3. Fill in the missing graphs or verbal description for each case consistent with the information given. Ignore
any points of abrupt change. Assume that any initial position or velocity not given is zero.
Verbal
Descripti
on
a)
b)
c)
An object is at
rest some
distance from the
origin, then it
moves with
constant velocity
further away
from the origin,
then it stops
A car is
moving with
constant
velocity, then
it accelerates
to a higher
constant
velocity.
Object speeds up
then it slows down
at the same rate
then it turns around
and speeds up
going backwards.
Finally is slows
down at it returns
to the origin.
x vs. t
v vs. t
a vs. t
a=0, except at
the instants when
it starts to move
or stops
d)
A mass
hanging
from a
spring is
moving
up and
down.
e)
An object starts
from rest, it
speeds up
uniformly, then it
moves with
constant
velocity, finally
it slows down
uniformly to rest.
f)
A ball is
tossed
upwards, it
rises in the
air and
returns to
the
ground.
4. A motion diagram is an illustration showing the velocity and acceleration vectors of a moving object.
Draw motion diagrams for cases (b), (c), & (e) in the problem above.
a) Draw enough instantaneous velocity vectors at equal time intervals to clearly illustrate each case.
b) Draw change in velocity vectors for each time interval and then draw acceleration vectors for each
time interval. Ignore the points of abrupt change.
Velocity (
Case (b):
) and acceleration (
) vectors (∆v vectors are directed the same as acceleration):
Case (c):
Case (e):
5. An equation of motion describes algebraically how the motion of an object depends on time. Write
equations of motion describing cases (d), (f), & (g) above [case (e) requires 3 sets of equations]:
Case (d): x(t) = A sin(t); v(t) = A cos(t) ; a(t) = -A sin(t),( where  =2π/T)
Case (f): x(t) = xo + vot –gt2/2 ; v(t) = vo -gt ; a(t) = -g (up is + and down is -)
Case (e) requires a separate set of formulas for each different acceleration:
From 0->t1: x(t) = a1t2/2; v(t) = a1t ; a(t) = a1
From t1-> t2: x(t) = x1 + vmax t; v(t) = vmax ; a(t) = 0
From t2-> t3: x(t) = x2 + vmax t - a2t2/2; v(t) = vmax -a2t ; a(t) = -a2
6. When an object changes its motion abruptly in a short amount of time (as we have seen in many
examples) we ignore the short time intervals for the sake of simplicity and also because these are negligible
compared to the longer intervals of motion. These abrupt changes often are drawn as a sharp angle or
broken line in the motion graphs. Explain why such abrupt changes are impossible in the real world.
An “abrupt change” implies a finite change in an infinitesimal amount of time, which is not possible.
7. There are two formulas that we can use to determine average velocity: vave=x/t or vave=(vi + vf)/2.
a) Which one of these formulas is always correct (by definition) and which formula is only correct
under certain conditions? What are the limitations of the “other” formula that is not always correct?
b) A car travels 12 miles using two different constant speeds. The car travels half the distance with a
speed of 10 mph and the other half distance with a speed of 30 mph. Determine the average speed here.
c) The car travels the same 12 miles back but this time the driver spends half the total time of the trip
traveling at 10 mph and the other half time traveling at 30 mph. Determine the average speed in the return
trip.
d) Sketch position vs. time graphs for both trips. Which trip took longer (b) or (c)?
e) If you changed your speed gradually from 10 mph to 30 mph during the 12 miles trip, which of the
two cases above would have the same average velocity?
f) Prove (or give a logical argument) that the “half-time” average velocity (c) will always be faster than
the “half-distance” average velocity (b).
a) The definition of average velocity (x/t) is always correct because it is the definition of average
velocity. The other formula is only correct for constant acceleration, or any other “equal-time” averages.
The important idea here is that there are different types of averages so you have to be careful with the
word average velocity.
b) 15 mph. Find the total time, then divide into the total distance.
c) 20 mph. First write formulas for the distances traveled at each speed in terms of the total time.
d) Trip (b) took longer (lower average speed). Graph is drawn in (f) below.
e) It would resemble (c) above.
f) There are many ways to prove this. I think the simplest one is to look at the graph of both cases.
Assume that the total distance is D and that it takes a time T for an equal-time average and a time T’ for
an equal-distance average. In the equal-distance average you are forced to spend more time traveling at
the lower speed and that reduces the overall average speed. So T<T”.
D
Higher speed
Lower speed
T/2
T
T’
8. A car (A) is waiting for the red light to change at an intersection. When the light changes the car
accelerates at a rate of 1 m/s2. At the moment that car A begins to move a second car (B) passes it going
at a constant speed of 8 m/s.
a) If neither car changes its motion, how soon will it take for car A to match the speed of car B? Which
car is ahead at that time?
b) If neither car changes its motion, how soon will it take for car A to overtake car B? Which car is
moving faster at that time? How far from the entrance of the intersection does this happen?
c) Draw graphs of position and velocity vs. time for the two cars. Use a single position graphs and a
single velocity graphs for both cars. Make sure your graphs agree with your answers to (a) and (b).
d) If car A stops accelerating after 12 seconds, when will it overtake B? At what location will this
happen?
a) 8 sec. Car B is ahead.
b) 16 sec and 128 m from start. Car A is faster.
c) x vs t
v vs t
16 s
8s
d) 18sec and 144 m.
9. A man is running to catch a bus at his top speed of 6 m/s. The man is 25 m behind the bus when the bus
starts to pull away from the curve with an acceleration of 1 m/s2.
a) Will the man catch the bus? If yes, find out when and where it happens. If no, determine how close he
got to the bus and the minimum constant speed he needed to catch it.
b) Draw a position graph that includes both the motion of the man and the bus in the same scale.
c) Repeat the problem assuming that the bus was accelerating at a rate of 0.5 m/s2.
. a) No. He comes 7 m short. He would need a minimum speed of 7.1 m/s to catch the bus.
b) x vs t
c) The man catches the bus at 5.37 s after running 32.2 m.
6s
10. Think about what happens when you toss a ball up into the air and it rises high above your head.
a) Roughly estimate the average acceleration of the ball while in the hand and compare to its
acceleration when it’s rising in the air. Which acceleration is higher in magnitude? Do they have the same
directions? Hint: Compare how far the ball travels in the hand to how far it travels in the air.
b) Assume the ball moves 1 meter upward while in the hand, and that it is 2 m above the ground when
it leaves your hand. In addition assume that the ball is moving with an initial speed upwards of 12 m/s
when it leaves your hand. Draw graphs of position, velocity, and acceleration vs. time for the motion of
the ball. Assume air resistance is negligible. Let g= 10 m/s2.
c) How long is the ball in contact with the hand? How long is the ball in free-fall?
d) How fast is the ball moving when it hits the ground?
e) Repeat the problem reversing the direction of + and - in your solution. Verify that this does not
change the value of the answers.
a) The acceleration in the hand is higher, since the ball rises higher than the armlength uses to accelerate
the ball. They have opposite directions.
b)
x vs. t
v vs t
a vs t
c) 0.17 s; 2.6 s; d) –13.6 m/s
11. We often ignore air resistance for the sake of simplicity. The air produces an acceleration that always
opposes the motion and, that in addition, depends on the speed of the object itself (the greater the speed the
greater the effect of air resistance). Consider the motion of the ball in free-fall in the problem above, but
now let’s consider the effect of air resistance in a qualitative way.
a) How would the shape of the graphs of motion in the problem above change as a result of air
resistance? Do not worry about numerical values; concentrate on the overall shape of the graphs. Explain
the changes you made.
b) How does the time going up for a given distance compare to the time coming down for the same
distance in the case of non-negligible air resistance? Justify your answer.
The basic effect is that the acceleration is no longer constant. Air resistance adds to “g” on the way up,
making the downward (–) acceleration greater, and it subtracts from “g” on the way down making the
downward (–) acceleration lower. The graphs are no longer symmetric since it takes longer to come down
than it does to go up. The maximum height is lower.
12. A balloon is rising with a constant speed of 5 m/s. It carries a basket with a person inside. When the
balloon is a 120 m above the ground the person holds an object outside the basket and lets it go.
a) Describe the motion of the object as seen by a person on the ground and as seen by the person in the
balloon’s basket.
b) What maximum height does the object rise above the ground?
c) How long does it take the object to reach its maximum height? ...to hit the ground?
d) How fast is it going when it hits the ground?
e) If the person had thrown the object downward with a speed of 10 m/s relative to the balloon (instead of
merely letting go), how long would the object take to hit the ground in that case?
a) The person on the ground would see the object go up first (because it would have the initial velocity of
the balloon 5 m/s) before it comes down. The person in the basket would see the object drop from the start.
b) 121.25 M;
c) 0.5 s; 5.4 s;
d) 49 m/s
e) The initial speed of the object is -5 m/s (it’s going down) relative to the ground so it takes only 4 s to
come down.
13. A ball is thrown straight upwards with a velocity vo. At the same time a ball is dropped from a height H
above the ground. Determine answers in terms of H, g, and vo.
a) At what time do the balls cross paths?
b) Suppose vo= 5 m/s and H= 6 m. Where would the balls cross paths?
c) In terms of vo and g, derive an expression for the maximum value of H for which the balls will cross
paths above the ground.
a) Write position formulas for each object. Since you want them to cross, set the positions equal to each
other and solve for the time. The solution is t=H/vo.
b) Plug-in numbers to formula above to find the time and then determine the position of the objects at that
time using position formulas. In this case you get negative position (-1.2 m), which indicate that the ball and
rock do not meet above the ground.
c) Set up the condition that the positions be positive…
14. A man is trying to lift a weight using various rope and pulley set-ups as illustrated below. Assume that
the man is pulling his end of the rope with a speed vo. This problem is not so important in this chapter but
will become more important in future chapters.
a) For each set-up determine the speed at which the weight rises. Explain your reasoning clearly.
b) If the man is accelerating his end of the rope at a rate "a", at what rate is the weight accelerating in
each case.
(i)
(ii)
(iii) vo
vo
vo
?
?
?
You want to revisit this problem when you begin to deal with masses and pulleys in future chapters. In
figure (a) the weight rises with the same speed and acceleration of the hand.
In figures (b) and (c) the weight rises at half the speed and acceleration of the hand. This is because any
length of string pulled wraps around the pulley and allows the pulley to move only half as much as the
amount of string that slides around it.
15. An elevator has a height of 3 m. While the elevator is moving a bolt inside the elevator drops from the
ceiling to the floor. Determine the time it takes for the bolt to hit the elevator floor in the following cases.
a) The elevator is moving up with a constant speed of 1 m/s.
b) The elevator is accelerating down at a rate of 2 m/s2 and the speed of the elevator is up 1 m/s at the
instant the bolt comes loose.
c) Describe the motion of the bolt in (b) as seen by an observer on the ground, and also as seen by an
observer inside the elevator.
d) Solve problem (b) using the frame of reference of the observer inside the elevator and show that the
answer is the same.
e) Repeat the problem assuming that the elevator is accelerating up at a rate of 2 m/s2 and the speed of the
elevator is up 1 m/s at the instant the bolt comes loose.
a) 0.77 ; b) 0.87 s
c) To an observer on the ground the bolt goes up first then comes down, all with the acceleration of “g=10
m/s2”. To the observer in the elevator, the bolt drops from rest with an acceleration 2 m/s 2 less than g,
inside the elevator the apparent acceleration of the bolt is 8 m/s2.
d) For the observer inside the elevator, the initial speed of the bolt is zero and the acceleration downward is
8 m/s2 (not 10 m/s2). Same answer as (b).
e) 0.71 s
16. The breaks are applied to a car traveling with speed vo and it comes to a stop in a distance "d". We
assume that the acceleration due to the breaks is constant for this car regardless of speed.
a) If the initial speed had been 2vo, how much longer would it take the car to stop?
b) In driving school they suggest that you allow one car length of stopping distance for every 10 mph of
speed of your car. Is this consistent with a constant deceleration from your brakes? Explain.
c) More realistically one has to allow for the "reaction time" that it takes before the foot hits the brakes.
During this time the car moves through a "reaction distance" at the initial constant velocity and this adds to
the "braking distance" to make up the total "stopping distance". The following table gives typical values:
Initial speed (m/s)
Reaction distance (m)
Braking distance (m)
Stopping distance (m)
10
7.5
5.0
12.5
20
15
20
35
30
22.5
45
67.5
What is the "reaction time" implied by the data in the chart above? What is the acceleration generated by
the brakes? Is the acceleration constant?
d) Determine the car’s stopping distance if the initial speed of the driver is 25 m/s?
a) It takes 4 times more distance and 2 times more time to stop with twice the initial velocity and same “a”.
b) No. The actual stopping distances increase in a squared fashion.
c) Reaction time =0.75 s. Break acceleration = -10 m/s2 and it is constant.
d) 50 m
17. A block slides over a rough patch of surface 40 cm long. When it enters the patch the speed of block is
12 cm/s and when it emerges its speed is 6 cm/s.
a) Determine the time the block spends moving over the rough patch.
b) Determine the acceleration of the block.
c) How much larger would the patch have to be in order to stop the block altogether (assume acceleration
is the same).
d) A second block enters the patch 1 s later moving with velocity 15 cm/s, will it overtake the first block
within the patch? Assume both blocks have the same acceleration moving over the rough patch.
a) 4.4 s; b) –1.35 cm/s2 ; c) 53.3 cm
d) Yes, the second block can catch up to the first within the 3.4 s it has to do so.
18. A train is traveling from station A to station B a distance D away. Starting at A the train first accelerates
uniformly to a maximum speed vmax then it decelerates uniformly to come to a stop at B. The train spends
twice as much time decelerating as it spends accelerating.
a) Draw graph of position, velocity and acceleration vs. time for the entire trip.
b) Determine the average speed during acceleration interval, during deceleration interval, and overall.
c) Determine the time for the complete trip in terms of vmax and D.
d) Determine the accelerations.
e) Would it have changed the overall time for the trip if the train had accelerated more quickly initially to
the maximum speed and then taken longer to come to a stop? Justify your answer.
a) vs. t
v vs t
a vs tv
b) v average= vmax/2 both in the accelerating phase and in the decelerating phase. So the overall average
velocity is vmax/2 for the entire trip. This is only true in this problem because the train never travels at a
constant speed of turn around.
c) 2D/ vmax
d) a1 =3 vmax 2/2D; a2 = -3 vmax 2/4D
e) No. The average velocity would be the same in any case.
Problems below require calculus to solve:
19. Review the proof of the kinematics formula for position under a constant acceleration [x = at2/2 + vot +
xo] which was done in class. Then proceed to proof the kinematics formula without time [2ax = v2-vo2].
Hint: Rewrite the definition of acceleration with the help of the chain rule as: a=dv/dt= (dv/dx)(dx/dt)]. Set
up the appropriate integral and solve.
Starting with definition of acceleration: a = dv/dt. Apply chain rule, a = (dv/dx)(dx/dt) = (dv/dx)(v). Break
up the differentials to set up an integral: a dx = v dv. Integrate both sides remembering that a is a constant
and you get (2a x) = v2 – vO2
20. Air resistance always opposes the motion and depends on the velocity of the object. It also depends on
some of the physical properties of the air (like its density) and of the object (such as its shape). If all the
physical properties that affect air resistance are combined into a single constant "C", the acceleration due to
air resistance can be written as a = -Cv2(i) (where i is the unit vector in the direction of the velocity).
a) Explain the meaning of the negative sign in this formula.
b) Derive a function of the velocity as a function of time due to air resistance on an object with initial
speed vo. Graph this function. Compare to the example of “fluid resistance” done in class (where a = -Cv).
c) How long would it take the object to slow down to half its initial velocity?…to come to rest?
d) The acceleration due to gravity can make an object "turn around" in free-fall. Can air resistance alone
make an object "turn around? Justify your answer.
a) The negative sign indicates that this acceleration is always in the opposite direction as the velocity.
b) Set up an integral using the fact that a=dv/dt. The answer: v = (Ct +1/vo)-1.
c) t=1/(Cvo). It takes an infinite amount of time to come to rest.
d) Air resistance always opposes motion so it cannot make an object turn around.
21. Galileo, who gave us our basic definition of acceleration, noticed that objects moving under the effect
of gravity traveled "odd multiples of distance in successive time intervals". This means that if an object falls
a distance "x" in a time "t" from rest , it would fall a distance "3x" during the next “t” interval and "5x" the
time interval after that…etc.
a) Show that this is consistent with our kinematics formula x=gt2/2 + vot.
b) Show that this is consistent with the meaning of a constant acceleration.
c) Galileo had considered defining acceleration as a=dv/dx instead of dv/dt. Derive a kinematics formula
for position vs. time for the case in which “a=dv/dx” is a constant.
a)Use the position formula to write and equation for (x2-x1), (x3-x2), etc. The displacements will be
proportional the odd numbers.
b) Start with the position formula and take two derivatives.
c) Start by writing an integral from the defninition of the accelertion and using the chain rule: a=dv/dx=
(dv/dt)(dt/dx)adt = dv/v. After integrating both sides you can derive the velocity vs. time function ( v =
voeat). Finally since v=dx/dt, you can set up another integral and derive the answer: x= xo + (vo/a)(eat -1)