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What are velocity vs. time graphs?
See what we can learn from graphs that relate velocity and time.
What does the vertical axis represent on a velocity graph?
The vertical axis represents the velocity of the object. This probably sounds obvious, but I have to warn you that
velocity graphs are notoriously difficult to interpret. People get so used to finding velocity by determining the
slope (as would be done with a position graph), they forget that for velocity graphs the value of the vertical axis is
giving the velocity.
Try sliding the orange dot horizontally on the example graph below to choose different times and see how the
velocity changes.
velocity (m/s)
5
4
3
2
1
1
2
3
4
5
6
7
time (s)
−2
−3
−4
−5
Concept check: What is the velocity of the object at time t
= 4 seconds according to the graph above?
[Show me the answer]
What does the slope represent on a velocity graph?
The slope of a velocity graph represents the acceleration of the object. So the value of the slope at a particular
time represents the acceleration of the object at that instant.
The slope of a velocity graph will be given by,
slope =
rise
v2 − v1
Δv
=
=
run
t2 − t1
Δt
dustin 


v (m/s)
v2
rise
run
v1
t1
t2
t (s)
Since
Δv
is the definition of acceleration, the slope of a velocity graph must equal the acceleration of the object.
Δt
slope = acceleration
This means that when the slope is steep, the object will be changing velocity rapidly. When the slope is shallow,
the object will not be changing its velocity as rapidly. This also means that if the slope is negative (directed
downwards), the acceleration will be negative, and if the slope is positive (directed upwards) the acceleration will
be positive.
Try sliding the orange dot horizontally on the example velocity graph below to see what the slope looks like for
particular moments in time.
velocity (m/s)
5
4
3
2
1
1
2
3
4
5
6
7
time (s)
−2
−3
−4
−5
The slope of the curve is positive between the times t
means the acceleration is positive.
The slope of the curve is negative between t
means the acceleration is negative.
= 0 s and t = 2 s since the slope is directed upward. This
= 2 s and t = 8 s since the slope is directed downward. This
At t = 2 s the slope is zero since the tangent line is horizontal. This means the acceleration is zero at that
moment.
Concept check: Is the object whose motion is described by the graph above speeding up, slowing down, or
traveling at constant velocity at time t = 4 s?
[Show me the answer]
What does the area under a velocity graph represent?
The area under a velocity graph represents the displacement of the object. To see why consider the following
graph of motion which shows an object maintaining a constant velocity of 6 m/s for a time of 5 s.
v (m/s)
7
6
5
4
3
2
1
1
2
3
4
5
t (s)
To find the displacement during this time interval we could use the formula,
Δx = vΔt = (6 m/s)(5 s) = 30 m
Which gives a displacement of 30 m.
Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of
area made by the graph as seen below.
v (m/s)
7
6
5
4
3
2
1
1
2
3
4
5
t (s)
The area of this rectangle can be found by multiplying height of the rectangle (6 m/s) times its width (5 s) which
would give,
area = height × width = 6 m/s × 5 s = 30 m
This is the same answer we got before for the displacement. The area under a velocity curve, regardless of the
shape will equal the displacement during that time interval. [Why is this still true when velocity isn't constant?]
area = displacement
What do solved examples involving velocity vs. time graphs look like?
Example 1: Windsurfing speed change
A windsurfer is traveling along a straight line and her motion is given by the velocity graph below.
Select all of the following statements that are true about the speed and acceleration of the windsurfer.
(A) speed is increasing
(B) acceleration is increasing
(C) speed is decreasing
(D) acceleration is decreasing
v (m/s)
7
6
5
4
3
2
1
1
2
3
4
5
6
7
8
9
t (s)
Options A (speed increasing) and D (acceleration decreasing) are both true.
The slope of a velocity graph is the acceleration. Since the slope of the curve is decreasing and becoming less
steep this means that the acceleration is also decreasing.
It might seem counter-intuitive, but the windsurfer is speeding up for this entire graph. The value of the graph,
which represents the velocity, is increasing for the entire motion shown but the amount of increase per second is
getting smaller. For the first 4.5 s the speed increased from 0 m/s to about 5 m/s, but for the second 4.5 s the
speed increased from 5 m/s to only about 7 m/s.
Example 2: Go-kart acceleration
The motion of a go-kart is shown by the velocity vs. time graph below.
a. What was the acceleration of the go-kart at time t
= 4 s?
= 0 s and t = 7 s?
b. What was the displacement of the go-kart between t
v (m/s)
7
6
5
4
3
2
1
1
2
3
4
5
6
7
a. Finding the acceleration of the go-kart at t
We can find the acceleration at t
slope =
t (s)
= 4 s:
= 4 s by finding the slope of the velocity graph at t = 4 s.
rise
run
For our two points we'll choose the start (3 s, 6 m/s) and end (7 s, 0 m/s) of the diagonal line as points 1 and 2
respectively. Plugging these points into the formula for slope we get,
slope =
v2 − v1
0 m/s − 6 m/s
−6 m/s
m
=
=
= −1.5 2
t2 − t1
7 s − 3 s
4 s
s
m
acceleration = −1.5
m
s2
b. Finding the displacement of the go-kart between t
= 0 s and t = 7 s:
We can find the displacement of the go-kart by finding the area under the velocity graph. The graph can be
thought of as being a rectangle (between t = 0 s and t = 3 s) and a triangle (between t = 3 s and t = 7 s).
Once we find the area of these shapes and add them, we will get the total displacement.
v (m/s)
7
6
5
4
3
2
1
1
2
3
4
5
6
7
t (s)
The area of the rectangle is found by,
area = h × w = 6 m/s × 3 s = 18 m
The area of the triangle is found by,
area =
1
1
bh = (4 s)(6 m/s) = 12 m
2
2
Adding these two areas together gives the total displacement.
total area = 18 m + 12 m = 30 m
total displacement = 30 m
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Top Recent
How would I find the displacement on the curved/changing acceleration charts? When the acceleration
is a straight line, I can use Base x Height for a rectangle, and (1/2)Base x Height for a triangle, but what
about the curves on the charts?
8 Votes
• Comment • Flag
5 months ago by
Edgar Sabin
You need calculus for that. That's why it was invented - to deal with continuous change.
• 17 Votes
• 4 Comments • Flag
5 months ago by
Andrew M
Show all 5 answers • Answer this question
How would you calculate the average speed on a velocity/time graph?
4 Votes
• Comment • Flag
4 months ago by
ayalaalyssa42
The area under the velocity/time curve is the total displacement. If you divide that by the change
in time, you will get the average velocity. Velocity is the vector form of speed. If velocity is always
non-negative, then average velocity and average speed are the same.
In example 2 above, the displacement of the go-kart was found to be 30m. The change in time
was 7 seconds, so the average velocity was 30/7 = 4.286 m/s. Since velocity was never negative,
this is also the average speed.