The slope of the curve is positive between the times t=0\text{ s}t=0 st, equals, 0, start a text, space, s, end text, and t=2 \text{ s}t=2 st, equals, 2, start a text, space, s, end text since the slope is directed upward. This means the acceleration is positive.
The slope of the curve is negative between t=2 \text{ s}t=2 st, equals, 2, start a text, space, s, end text, and t=8 \text{ s}t=8 st, equals, 8, start a text, space, s, end text since the slope is directed downward. This means the acceleration is negative.
At t=2\text{ s}t=2 st, equals, 2, start a text, space, s, end text, 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 or slowing down at time t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text?
[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 to consider the following graph of motion that shows an object maintaining a constant velocity of 6 meters per second for a time of 5 seconds.
To find the displacement during this time interval, we could use this formula
\Delta x=v\Delta t=(6\text{ m/s})(5\text{ s})=30\text{ m}Δx=vΔt=(6 m/s)(5 s)=30 mdelta, x, equals, v, delta, t, equals, left parenthesis, 6, start text, space, m, slash, s, end text, right parenthesis, left parenthesis, 5, start text, space, s, end text, right parenthesis, equals, 30, start text, space, m, end text
which gives a displacement of 30\text{ m}30 m30, start text, space, m, end text.
Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of the area made by the graph as seen below.
The area of this rectangle can be found by multiplying the height of the rectangle, 6 m/s, times its width, 5 s, which would give
\text{ area}=\text{height} \times \text{width} = 6\text{ m/s} \times 5\text{ s}=30\text{ m} area=height×width=6 m/s×5 s=30 mstart text, space, a, r, e, a, end text, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text, equals, 6, start text, space, m, slash, s, end text, times, 5, start text, space, s, end text, equals, 30, start text, space, m, end text
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?]
\text{area under curve}=\text{displacement}area under curve=displacementstart text, a, r, e, a, space, u, n, d, e, r, space, c, u, r, v, e, end text, equals, start text, d, i, s, p, l, a, c, e, m, e, n, t, end text
[Wait, aren't areas always positive? What if the curve lies below the time axis?]
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.
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 counterintuitive, 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 seconds, the speed increased from 0 m/s to about 5 m/s, but for the second 4.5 seconds, 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\text{ s}t=4 st, equals, 4, start a text, space, s, end text?
B. What was the displacement of the go-kart between t=0\text{ s}t=0 st, equals, 0, start text, space, s, end text and t=7\text{ s}t=7 st, equals, 7, start text, space, s, end text?
A. Finding the acceleration of the go-kart at t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text
We can find the acceleration at t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text by finding the slope of the velocity graph at t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text.
\text{slope}=\dfrac{\text{rise}}{\text{run}}slope= Â
run
