Name
Period
Date
Uniform Motion Review
Topics:
 Multiple representations of Uniform Motion:
o Position vs. time and velocity vs. time graphs
o Verbal descriptions of motion
o Motion Maps
o Mathematical Models (i.e. developing and using equations from graphs as
practiced in the Scientific Reasoning unit and the Uniform Motion lab).
 The relationships between different methods of representing uniform motion (given a
position graph, you must be able to produce a velocity graph or motion map etc.).
 Uniform Motion Equations (in one dimension):
s=
distance traveled
time interval
v=
Dd
Dt
(these equations
provided)
 Dd = change in position = displacement
 The difference between distance, displacement, speed, and velocity
 Calculating the velocity from a position vs. time graph. Also determining the physical
interpretation of slopes & intercepts of position vs. time and velocity vs. time graphs.
Relevant Material to Study:
 All Handouts, Worksheets, and Tutorials
 Results from the Uniform Motion Lab
 Uniform Motion Lab Practicum
 Chapter 2 in the Textbook up to the Acceleration sections
Note: to ensure you are well prepared for the test, you must be able to complete the following problems as if they
were the test. That means no help, no notes, no friends, no extra time etc. If you cannot understand these
problems the first time you attempt them, you are not prepared! Finally, when you review the solutions pay
close attention to details like units and significant figures. Most students lose as many points for these details as
they do for not understanding the Physics.
Part I: Timed Practice Problems (50 minutes)
1. Consider the position vs. time graph at right describing the
motion of an African gazelle.
a. Determine the physical interpretation of the vertical
intercept.
The vertical intercepts represents the value of the position
when the time is zero. Alternatively: starting position.
b. Determine the average velocity of the gazelle. Make sure to
consider significant figures.
To determine velocity from a position vs. time graph, we must determine the slope of
the line.
60. 𝑚 − 10. 𝑚
≈ 8.3 𝑚/𝑠
6.0𝑠 − 0.0𝑠
c. Write a mathematical equation to describe the motion of the gazelle. Make sure to
include units and carefully consider the number of significant figures you record.
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = 8.3
𝑚
(𝑇𝑖𝑚𝑒) + 10. 𝑚
𝑠
d. Based on the provided data, what can you determine about how fast an African gazelle will
tire?
Since the slope of the graph is linear between t=0s and t=6s, then we know the gazelle’s
velocity is constant in this region. Therefore based on this data, we conclude that
gazelles do not tire in less than six seconds.
e. Assuming the gazelle does not tire, where will the gazelle be after 9s?
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = 8.3
𝑚
(9.0𝑠) + 10. 𝑚 ≈ 83 𝑚
𝑠
2. Shown at right is a velocity vs time graph for an object.
Positive velocities represent north while negative south.
a. Completely describe the motion of the object.
Our object starts at an unknown location. From this point, it travels north at a constant
velocity of 4 m/s for 2 seconds. Then the object turns
around and travels for 1 second at 3 m/s south. Finally
the object rests for 2 seconds.
b. Sketch (no numbers) the corresponding position vs time
graph.
c. Suppose the object is traveling at 4 m/s for the first 2s of
its motion. What was its displacement during the time
t=0s to t=2s?
𝑚
∆𝑑 = 𝑣⃑∆𝑡 = 4 𝑠 (2𝑠 − 0𝑠) = 8𝑚
8m north
3. Johnny drives north to Wisconsin (1920 miles) in 32 hours. He returns home by the same
route in the same amount of time.
a.
Determine the displacement between his home and Wisconsin one way, starting at
his home and ending in Wisconsin.
1920 miles north
b.
Determine his average speed.
c.
Speed = Total Time = 32Hr+32Hr = 64.Hr = 60. mi/hr
Determine his average velocity.
He had no displacement since he ends where he started.
𝑣𝑎𝑣𝑔 =
∆𝑑
0 𝑚𝑖
=
= 0 𝑚𝑖/𝐻𝑟
∆𝑡
64 𝐻𝑟
d.
Compare these two values and explain any differences.
Total Distance
1920mi+1920mi
3840mi
Society often uses the words “speed” and “velocity” interchangeably.
However, as Physicists, we need to distinguish between quantities like
velocity that have direction and those that do not like speed. Since these two
words have different definitions, it shouldn’t be surpassing they have
different values. To be successful, you need to learn the very precise
mathematical definition of the limited number of words we will define.
4. A basketball initially travels at 3 meters per second for 3 seconds:
a. Determine the velocity of basketball at t=0s.
The vertical intercept of a velocity vs. time graph represents
the starting velocity. In this case vi=3 m/s.
b. Determine the physical interpretation of the slope of a velocity
vs. time graph. Note: we have not discussed this yet, but think
about a general method for interpreting slopes –learned in the
Scientific Reasoning unit – and you can figure it out!
𝑚= 0
𝑚
𝑠
1
𝑠
The slope of a velocity vs. time graph represents the change in velocity for each
additional unit time. For now, velocities are not slowly changing over time (i.e. that’s what
it means to have a constant velocity!), so every slope of a velocity vs. time graph has been:
So far, we have only seen horizontal lines on velocity vs. time graphs.
c. Describe the motion of the ball after t = 3 seconds.
At t=3s the ball moves in the negative direction at a constant speed of 1 m/s for 4s.
d. Draw a quantitative motion map that models the motion of the object. Remember to label
velocity vectors and the t=0s dot.
5. A race car travels at a speed of 100. m/s. How far does it travel in 12.5 s? Use the appropriate
mathematical expression and show how units cancel. (Keep the proper number of sf's.)
⃑⃑ =
𝑣
∆𝑑
∆𝑑
∆𝑡
= 𝑣⃑∆𝑡 = 100.
𝑚
𝑠
∗ 12.5𝑠 = 1250 𝑚
6. Below is a qualitative motion map for Wandering Willie. On the coordinate axes below
sketch a graph which generally describes Willie's motion (you need not plot points).
⃑⃑
𝑣
t=0
7. A bird travels toward the origin, then suddenly reverses direction. Make sure to consider
significant figures for each question below.
a. Find the average velocity from
t = 30s to t = 40s.
10. 𝑚 − 30. 𝑚 −20. 𝑚
=
40. 𝑠 − 30. 𝑠
10. 𝑠
= −2.0 𝑚/𝑠
⃑⃑ =
𝑣
2.0 m/s towards the origin.
b. Find the average velocity from
t = 40s to t = 50s.
⃑⃑ =
𝑣
1.5 m/s away from the origin.
c. Determine the average speed from t = 30s to t = 55s.
𝑠𝑝𝑒𝑒𝑑 =
𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
Total distance = 20m + 15m = 35m
Total time = 25s
𝑠𝑝𝑒𝑒𝑑 =
35𝑚
= 1.4 𝑚/𝑠
25𝑠
25. 𝑚 − 10. 𝑚 15. 𝑚
=
= 1.5 𝑚/𝑠
50. 𝑠 − 40. 𝑠
10. 𝑠
d. Determine the average velocity from t = 30s to t = 55s.
⃑⃑ =
𝑣
⃑⃑⃑⃑⃑⃑
∆𝑑
∆𝑡
=
25𝑚 − 30. 𝑚
= −.20 𝑚/𝑠
55𝑠 − 30. 𝑠
.20 m/s towards the origin.
e. Find the velocity at t = 35 seconds.
We need to determine the slope of the graph at t=35s to determine the velocity at that
instant. Since the slope at t=35s is linear then:
⃑⃑ =
𝑣
10. 𝑚 − 30. 𝑚 −20. 𝑚
=
= −2.0 𝑚/𝑠
40. 𝑠 − 30. 𝑠
10. 𝑠
2.0 m/s towards the origin.
x vs t
A
B
C
X(m)
D
E
t(s)
8A. Rank the speeds of the five motorists, from the position vs. time graph above.
greatest
A
B
D
E
C
least
Explain your reasoning.
First, it is important to note we were asked to rank speed, not velocity. This means that we
ignore all directions. We can “see” the velocity of an object by examining its slope on a
position vs. time graph, with steeper slopes representing greater speeds. Consequently one
needs only rank the slope steepness. It’s clear that A’s slope is steeper then B’s. B’s slope is
steeper then D’s since B’s slope is about 2 m/s. D is more like -1 m/s. E is clearly less steep
then D, and C has a slope of 0 m/s.
8B. Explain major differences in the motion of motorists A & E.
There are several important differences between the motions A & E. First, A starts at the
origin while E starts with some positive position. Second, A has a much great speed (see
previous question). Finally, A and E are moving in opposite directions. We can see that A’s
line has a positive slope – indicating it is moving away from the origin. However, E has a
negative slope which implies it is moving towards the origin.
9. Which of the following situations represents a negative displacement? (Assume positive
position is measured vertically upward along a y-axis.) (check all that apply)
__A. A cat stands on a tree limb.
__B. A cat jumps from the ground onto a tree limb.
__C. A cat jumps from a lower tree limb to a higher one.
X_D. A cat jumps from a tree limb to the ground.
__E. None of the above
10. Which of the following line segments on a velocity versus time graph is physically
impossible?
a. horizontal line
b. straight line with negative slope
c. straight line with positive slope
d. vertical line
A vertical line would imply that an object is at multiple positions at the same instant
in time. This is unreasonable.
Part II: Extra Practice
1. A student runs 150 m due north in 20 s and then jogs 30 m due south in 10 s. Finally the
student doubles back 5m north again in 5 s.
Find the following (you do not need to show your work for this problem):
A. Total distance that the student traveled = 150m + 30m + 5m = 185m
B. Total displacement of the student = Let north be positive and south negative.
150m + -30m + 5m = 125m. Thus the student’s displacement is 125m north.
C. The student’s average speed = 5.3 m/s
D. The student’s average velocity = 3.6 m/s north
2. Consider the v vs t graph right.
a. Completely describe the behavior of the object depicted in the graph.
The object starts at some unknown position. The object then moves at a
large speed in the positive direction for a period of time at a constant
velocity. The object then turns around and heads in the negative
direction at a slower constant speed for a longer time.
b. Draw a motion map that represents the behavior of the object.
Remember to label t=0s and velocity vectors.
3. Compare the velocities of the cyclists, A and B, at the time t= 5.0s on the position-time graph
right.
B>A
Explain your reasoning.
To determine velocity, we need to look at the slope of
the lines corresponding to motions A and B. It’s clear
that A’s slope is horizontal so it is not moving. However,
B has a positive slope at t=5.0s, so it must have a greater
velocity.
4. Jimmy makes the following map diagram for the graph right.
Find any error(s) and correct them. Explain what, if anything, is wrong with Jimmy’s motion
map, and how you corrected the error(s).
Jimmy’s motion map for object B looks fairly good. He recognized the object starts at the
origin (although he should have labeled that) and moves with a constant velocity away
from the origin. He should have added a t=0s on the first dot though.
His motion map for A is incorrect. Object A should be at rest since its slope is horizontal.
On the motion map, Jimmy should have drawn dots stacking up since their positions are all
the same. Additionally object A starts with some positive position. Unfortunately Jimmy
did not show this on the motion map by starting the t=0s for object at a positive position. A
corrected motion map is drawn below. Notice how A and B are at the same position at t=3s.
5. Consider the following two equations:
Position of cyclist A = 94 meters
3.8 m/s x 30.0 s = 114 m (position of cyclist B)
Describe a physical situation for which each of the equations above applies (i.e. tell a story):
Carefully explain your reasoning.
Once upon a time there were two cyclists Randall (cyclist A) and Shirley (cyclist B). On one
fine morning, Randall took a leisurely ride along Saratoga Ave. He biked past Burger King
and the Casino while taking note of the glorious shrews and birds about. However his
morning was not destined to be pleasant. For no sooner then had he passed the San Tomas
Expressway when tragedy stuck: his tire went flat. Not knowing what to do, he sat in
despair for several hours.
Finally, in his despair, he resolved to call his mechanic-friend Shirley who also owned a
bike. Upon learning of his misfortune, she quickly agreed to come to his aid. In fact, Shirley
lived along Saratoga Ave, about 20 meters beyond where Randall had started his ride.
Shirley peddled at her top speed of 3.8 m/s taking 30s to reach her friend. Once there, she
expertly fixed his tire and they both continued on a winding journey – living happily ever
after.
In order for my story to work, the two cyclists had to meet up. Since cyclist B traveled
114m and cyclist A starts 94m from some arbitrary point then it was necessary that cyclist
B start 20m behind where cyclist A did. Since cyclists A position is constant there needed to
be a reason for him to remain in place (i.e. the flat tire). 3.8 m/s seemed like a reasonable
top speed on a bike for the second cyclist to ride to the first’s aid.