How to analyze graphs that relate velocity and time to acceleration and displacement. 

What does the vertical axis represent on a velocity graph?

The vertical axis represents the velocity of the object. This probably sounds obvious, but be forewarned—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 dot horizontally on the example graph below to choose different times and see how the velocity changes.
Concept check: What is the velocity of the object at time t=4 secondst=4\text{ seconds}, according to the graph above?
The velocity is +3 m/s+3\text{ m/s} at time t=4 st=4\text{ s}. We can see this by moving the dot to the time t=4 st=4\text{ s} and seeing that the value of the graph is +3 m/s+3\text{ m/s} on the vertical axis.

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 the following formula:
slope=riserun=v2v1t2t1=ΔvΔt\text{slope}=\dfrac{\text{rise}}{\text{run}}=\dfrac{v_2-v_1}{t_2-t_1}=\dfrac{\Delta v}{\Delta t}
Since ΔvΔt\dfrac{\Delta v}{\Delta t} is the definition of acceleration, the slope of a velocity graph must equal the acceleration of the object.
slope=acceleration \text{slope}=\text{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 dot horizontally on the example velocity graph below to see what the slope looks like for particular moments in time.
The slope of the curve is positive between the times t=0 st=0\text{ s} and t=2 st=2 \text{ s} since the slope is directed upward. This means the acceleration is positive.
The slope of the curve is negative between t=2 st=2 \text{ s} and t=8 st=8 \text{ s} since the slope is directed downward. This means the acceleration is negative.
At t=2 st=2\text{ s}, the slope is zero since the tangent line is horizontal. This means the acceleration is zero at that moment.
Concept check: Is the velocity of the object whose motion is described by the graph above increasing, decreasing, or constant at time t=4 st=4\text{ s}?
The object's velocity is decreasing at time t=4 st=4\text{ s}. One way to see this is by dragging the slider to t=4 st=4\text{ s}. Notice that the slope of the graph is negative, which means the acceleration is negative. The velocity at t=4 st=4\text{ s} is positive and has a value of +3 m/s+3\text{ m/s}. Since the acceleration and the velocity have opposite signs, the object must be slowing down.
Another way to see this is by recognizing that at t=4 st=4\text{ s} the curve is approaching the horizontal axis. Since the horizontal axis represents v=0 m/sv=0\text{ m/s}, a curve that is heading toward the horizontal axis represents an object that is slowing down. Similarly, a curve that is heading away from the horizontal axis represents an object that is speeding up. So at t=7 st=7\text{ s} in the graph above, the curve is heading away from the horizontal axis and the object's velocity is increasing in the negative direction.

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 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
Δx=vΔt=(6 m/s)(5 s)=30 m\Delta x=v\Delta t=(6\text{ m/s})(5\text{ s})=30\text{ m}
which gives a displacement of 30 m30\text{ 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.
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\text{ area}=\text{height} \times \text{width} = 6\text{ m/s} \times 5\text{ s}=30\text{ 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.
For an arbitrary shape, you can consider breaking the area into rectangles of very small width as seen in the graph below. The displacement during each small time interval would be given by Δx=vΔt\Delta x=v\Delta t since the velocity would be roughly constant if the time intervals are small enough. If you were to make the rectangles infinitesimally small, they would exactly represent the area under the curve. Adding up all the small areas would equal the total area and also the displacement of the object. When doing an actual problem, you never have to actually break a curve into an infinite amount of rectangles. Once you have proven to yourself that the area represents the displacement, you are free to determine the area in whichever way is most convenient.
area=displacement\text{area}=\text{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.
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 st=4\text{ s}?
B. What was the displacement of the go-kart between t=0 st=0\text{ s} and t=7 st=7\text{ s}?

A. Finding the acceleration of the go-kart at t=4 st=4\text{ s}

We can find the acceleration at t=4 st=4\text{ s} by finding the slope of the velocity graph at t=4 st=4\text{ s}.
slope=riserun\text{slope}=\dfrac{\text{rise}}{\text{run}}
For our two points, we'll choose the start—3 s,6 m/s3\text{ s}, 6\text{ m/s}—and end—7 s,0 m/s7\text{ s}, 0\text{ m/s}—of the diagonal line as points one and two respectively. Plugging these points into the formula for slope we get
slope=v2v1t2t1=0 m/s6 m/s7 s3 s=6 m/s4 s=1.5ms2\text{slope}=\dfrac{v_2-v_1}{t_2-t_1}=\dfrac{0\text{ m/s}-6\text{ m/s}}{7\text{ s}-3\text{ s}}=\dfrac{-6\text{ m/s}}{4\text{ s}}=-1.5\dfrac{\text{m}}{\text{s}^2}
acceleration=1.5ms2\text{acceleration}=-1.5\dfrac{\text{m}}{\text{s}^2}

B. Finding the displacement of the go-kart between t=0 st=0\text{ s} and t=7 st=7\text{ 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 st=0\text{ s} and t=3 st=3\text{ s}) and a triangle (between t=3 st=3\text{ s} and t=7 st=7\text{ s}). Once we find the area of these shapes and add them, we will get the total displacement.
The area of the rectangle is found by
area=h×w=6 m/s×3 s=18 m\text{area}=h\times w=6\text{ m/s} \times 3\text{ s}=18\text{ m}
The area of the triangle is found by
area=12bh=12(4 s)(6 m/s)=12 m\text{area}=\dfrac{1}{2} bh=\dfrac{1}{2} (4\text{ s})(6\text{ m/s})=12\text{ m}
Adding these two areas together gives the total displacement.
total area=18 m+12 m=30 m\text{total~area}=18\text{ m}+12\text{ m}=30\text{ m}
total displacement=30 m\text{total~displacement}=30\text{ m}
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