Some practice with velocity and position graphs
Position to Velocity
The main idea here is that the velocity is the rate of change of the position. A large velocity means the
position changes fast, a big change (up/down on the graph) in a little time(sideways on the graph), making a
large slope on the graph. A small velocity means that there is very little change in position (up/down on the
graph) in the same amount of time, making a small slope on the graph. So the velocity is shown by the slope
of the line on the position graph (slope = rate of change).
Thus to find velocity we look for the rise (∆x) divided by the run (∆t), giving us the velocity in m/s.
For a graph with constant slopes this is fairly easy. The velocity is the same for all the points which have the
same slope, making a set of ‘flat lines’:
N
4
3
2
1
0
-1
-2
-3
-4
S
Each segment on the position graph gave a straight line on the velocity graph, with value = to slope.
More complex curves can be thought of as made of line segments. Try this one:
E6
5
4
3
2
1
0
1
2
3
W4
Of course a REAL position time graph does not have these sudden changes in velocity, it would be a smooth
curve. Finding slopes for a curve is a bit more difficult, this is described in the next section!
Finding the slopes of a curve:
Although an object may be changing its velocity, at any instant of time the object is moving with only one
speed and in only one direction! Finding that speed and direction from a position vs. time graph means we
have to find the slope of a curve at a point.
Here the idea is that curves can be approximated (faked, sort of) as a series of short straight lines – up close
the curve is “almost” a line. Thus the closer one comes to an actual point the more a curve looks like a
straight line. To see this imagine “zooming in” on a curve as shown below:
The closer in you
zoom, the more the
curve looks like a
straight line.
The slope of the curve at a point is the slope of the “straight line” that the curve resembles/becomes.
To find this line you could use calculus, which was designed by physicists to do this sort of thing, or you
can use your own very powerful visual processor – your brain! You are VERY good at seeing lines, so just
free your processor to do that by hiding everything except a small segment of the line (I use my thumbs and
a ruler like this:
we want to find
the slope here
Use thumbs to hide
the line except near
the point. Position the
ruler to match the line
segment.
Use ruler to draw a
line extending the line
segment so that its
slope can be found
The slope of the line
you drew is the slope
of the curve at that
point!
You know the line you drew is just right if you can’t tell the difference between the line and the curve right
at the point you are using! This line is called the tangent line.
You don’t need to find the tangent line at every point, just a few so you can connect them with a smooth
line. Some things to watch for:
• Where is the slope zero (flat).
• When is it up or down (east or west, + or -, …)?
• What is the steepest slope and what is the least steep?
• Where is the slope (velocity) roughly constant? Where is it changing quickly (very curved)?
You can see an example of this next!
An Example Using a Curved Position-Time Graph
These methods allow us to find approximate sketches quickly, or to make more detailed measurements and
make a pretty good quality graph of velocity.
v
constant v
v
v=0
v
changing v
make some tangents
N5
4
3
2
1
0
1
2
3
4
S5
Some things to watch for:
• It is common to see velocity as more constant than it really is!
• A parabola on a position-time graph produces a straight line on a velocity-time graph.
• The amount of curve in the position-time graph tells you about the way the velocity-time graph will
angle, more curve means more rapidly changing velocity which means a steeper slope on the
velocity graph.
Next we’ll go backwards: from velocity to position.
-2
Velocity to Position
If velocity is constant or we know the average velocity then getting the position is easy: we find the change
in position from the velocity and time, and add up changes if necessary.
∆s = vave ∆t
Adding up the changes in position as time passes gives us the overall change in position up to that point.
Eg: A person walks jogs east at 2.0 m/s for 2.0 seconds, then east at 3.0m/s for 4.0s, stops for 3.0s and
finally sprints east for 1.0s at 4.0m/s.
On a velocity graph that looks like this:
Motion East and West
Times
0-2s
2-6s
6-9s
9-10s
∆t
2.0s
4.0s
3.0s
1.0s
vave
2.0m/s E
3.0m/s W
0.0m/s
4.0m/s E
∆s
4.0 m E
12 m W
0m
4.0 m E
Position(0 m)
4.0 m E
8.0 m W
8.0 m W
4.0 m W
Motion East and West
The positions we found are the positions at the END of
each time interval, so we put them into our graph of
position vs. time at these times.
In this case the velocity is constant in between the steps,
and we recall that a constant velocity on a position graph
is a straight line segment. So that is how we link the
points. When we get to curves it will be a little fancier.
As we saw in class, a graph where velocity changes suddenly between constant values is not very realistic.
However if the graph is a straight line, or even a curve, we can “turn it into” such a graph by using the idea
of average (effective) velocity. Consider something changing its velocity in a regular way:
Motion with constant changes
Motion with constant changes
First we “chop up” the time into intervals. I have chosen 0-3s, 3-5s, 5-7s, 7-10s. You could choose others,
but too few and you don’t get enough of a feel, too many and you do a lot of work you probably don’t need
to. I specifically choose to make one “chop” at 7.0s because that is where the velocity is 0 (a turn-around
point). Now I find an average velocity for each region and a distance travelled:
Times
veff
total ∆x
∆t
∆s
Initial
0 m (assumed)
Position at 3s
0-3s
3.0s
4.0m/s E
12.0m E
12.0 m E
Position at 5s
3-5s
2.0s
2.1m/s E
4.2 m E
16.2 m E
5-7s
2.0s
0.8m/s E
1.6 m E
17.8 m E
Etc.
7-10s
3.0s
1.1 m/s W
3.3 m W
14.5 m E
With curves we do the same thing, but making the “horizontal line” is a bit trickier. The secret is to get the
same amount above and below the graph:
5
5
5
5
4
4
4
4
3
3
3
3
Chop the time intervals:
key points: max/min
points, zero points.
Place average velocity.
This is too high, the
upper part is smaller
than the lower.
This is too low, the
lower part is smaller
than the upper.
This is just right, the
two parts are equal.
So vave is read from the
vertical axis ≈ 3.7m/s
I call this the “bow tie” method, because getting the two halves equal is like making a bow tie work.
Some practice problems:
Questions 1 to 3 refer to the following VELOCITY vs. TIME graphs:
a) +3
b) +3
c) +3
v(m/s 0
)
t(s)
5
v(m/s
)
-3
t(s)
v(m/s 0
)
5
-3
d)
t(s)
5
+3
v(m/s 0
)
-3
t(s)
5
-3
1. Which of the graphs shows the largest displacement at the end of the time shown?
2. Which graph shows a constant velocity?
3. Which graph shows an object that changes direction during the time shown?
Questions 4 to 6 refer to the POSITION-TIME graphs below: (notice that these show something quite
different than the first set (from questions 1 to 3)
a) +3
b) +3
c) +3
d) +3
x(m) 0
t(s)
5
x(m) 0
-3
t(s)
x(m) 0
5
-3
t(s)
5
x(m) 0
-3
t(s)
5
-3
4. Which graph or graphs show a constant velocity?
5. Which graph would have the greatest displacement at the end of the time shown?
6. Which graph shows an object that changes direction in the time shown?
Questions 4 to 8 refer to the following DISPLACEMENT vs. TIME graph for a moving car.
x (km)
Position of Car
30.0E
20.0E
10.0E
0
10.0W
20.0W
30.0W
0 1 2 3 4 5 6 7 8 9 10 11
t (hr)
7. At which of the following times does the car pass the (arbitrary) origin?
a) 0 hr
b) 2 hr
c) 7hr
d) 9hr
8. When does the car have zero velocity?
a) 0 hr
b) 2 hr
c) 7hr
d) 9hr
9. What is the approximate size of the average velocity of the car between 0 hr and 7 hr?
a) 0 km/hr
b) 3 km/hr
c) 4 km/hr
d) 7 km/hr
10. What is the size of the instantaneous velocity at 5 s?
a) 0 km/hr
b) 1 km/hr
c) 7 km/hr
d) 13 km/hr
11. What is the car’s total displacement in the time shown?
a) 10 km
b) 20 km
c) 26 km
d) 40 km
12. Between what times is the car going west?
a) 0 to 7 s
b) 0 to 4.7s
c) 4.7s to 9s
d) after 7 s
Answers:
1.
7.
2.
8.
3.
9.
4.
10.
5.
11.
6.
12.
e) never
e) never
e) 10 km/hr
e) 30 km/hr
e) 90 km
e) after 9 s
Solving Word Problems (extracting information)
Examine the following word problem. Read it over, and try to imagine the motion.
A car is stopped at a red light on Granville at No.2 Road. After 5.0 seconds the light turns
green and the car moves east at 12 m/s for the next 15.0 seconds. What is the car’s
average velocity over this entire time?
1) In the problem above, circle the words that tell you about the cars motion. For example the fact that the
car is stopped tells you about it's motion, so you should circle that word. There are others, some
describing position, some velocity, some time.
2) Now, next to each circle you made in question 1, write the variable symbol for the quality that it tells
v
you about. For example the "stopped" tells you about a velocity, so write v next to that circle.
3) If there is more than one different velocity then give the velocities different names so you can tell them
v
v
apart (like v1 and v2). Add the names to the variables you wrote beside the problems
4) Now write down the velocity or velocities in a list, including direction. Use the space below. Make sure
you include all the circled velocity information.
INFO:
v
∆time1
v1 =
v
v2
=
∆time2
FIND:
5) Write down the times that each of the velocities applies. Use the circled time information, and write the
starting and ending time that the velocity applies
6) Draw a picture showing the car stopped at the light. Add another image of the car, showing where it
moves (don't worry about numbers yet… just a general idea). This is to help you be clearer about what is
happening.
7) Add what you put into your list to the diagram.
8) Add what we are trying to find to the list above. Think about what that term means, and write down how
one finds it (in this case this might be a formula. (If there is more than one way to find it then you might
write down more than one thing, though at the moment I think there is just one definition for this
problem). Make a note of what you need to know to find it.
9) You should see that you need to know the change in position over the whole time and the change in
time. Now let's figure out how far the car has gone. Look at the first velocity… given how long the car
has this velocity, how far will it go in that time?
. How long does the car have the second
velocity in the problem?
In that time how far will it go?
. What is the total
distance the car goes?
What is the total time for the problem?
10) You are pretty much done now! Write down the answer clearly.
Here is my version of doing this problem:
A car is stopped at a red light on Granville at No.2 Road. After 5.0 seconds the light turns
green and the car moves east at 12 m/s for the next 15.0 seconds. What is the car’s
average velocity over this entire time?
v1=0,
INFO:
v
v1 =
5.0s
v2=12 m/s E,
15.0s
∆s
SOL:
CALC:
v
v2
=
FIND:
0 m/s
∆time1 =
5.0 s
12 m/s east
∆time2 =
15.0 s
vave
Find ∆s 1 = v1∆t1, and the same for ∆s 2. Find total ∆s , ∆t. vave= ∆s /∆t
∆s 1= 0m, ∆s 2= (12 m/s E)(15.0s) = 18|0 m E, so ∆s total = 18|0 m E
∆ttotal = 5.0s + 15.0s = 20.0 s
vave= 18|0 m E / 20.0s = 9.0 m/s E
Some word problems:
1. If you travel 150m N in 2.0 minutes, what is your effective velocity in m/s? 1.25 m/s N (1.3 m/s N)
2. If you run at 8.0 m/s for 3.0 minutes and then jog at 2.5 m/s for another 2 minutes, how far have you
gone?
1.7 km
3. What was your average (effective) velocity over the full time in the previous question?
5.8 m/s
4. You travelled at 30 km/h for 12.50 minutes to get to Richmond centre, which is 1.50 km E of your
reference point (your house). What was your initial position you must have begun travelling at?
Either we started at 4.75 km W (going 6.25 km E) or we started at 7.75 km E (going 6.25 km E)
5. You are running a race and travelling at 7.89 m/s from your coach's perspective. There is only 1 person
ahead of you! They are 50. m ahead to be exact. You catch up to the person in 25.3 s. What is the
velocity of the person you caught up to, as measured by your coach, who is standing still watching the
race from the sidelines?
5.91 m/s
6. Ms. Seminutin and I are going to run a race. Ms. Seminutin is much faster than me, since he can run at
8.5 m/s while I can only run at 6.2 m/s. If we are running a 200.0 m race how long a head start should
Ms. Seminutin give me to make the race fair (to make us cross the finish line at close to the same time)?
54m or 8.7s
7. If Ms. Seminutin decided to give me a head start of 6.0 seconds in the previous question, at what time
would he pass me?
22 s after I start
8. How far from the finish line would we be?
63 m
9. A car is speeding down the highway at 140 km/h when it passes a police car. The police car takes about
1.0 minutes to get started, but then travels at 160 km/h. How long does it take the police car to catch up
to the speeder?
3.5 min
10. You are heading to the school, which is 500m north of where you are. You walk quite quickly, at about
1.2 m/s. Someone else is heading south away from the school at the same time. If you meet them when
you are 150m from the school, what was their velocity?
0.51 m/s S