What are acceleration vs. time graphs?

See what we can learn from graphs that relate acceleration and time.

What does the vertical axis represent on an acceleration graph?

The vertical axis represents the acceleration of the object.
For example, if you read the value of the graph shown below at a particular time, you will get the acceleration of the object in meters per second squared for that moment.
Try sliding the dot horizontally on the graph below to choose different times, and see how the acceleration—abbreviated Acc—changes.
Concept check: According to the graph above, what is the acceleration at time t=4 st=4\text{ s}?
At t=4 st=4\text{ s} the acceleration is +1ms2+1\dfrac{\text m}{\text s^2}. We can verify this by sliding the dot horizontally and seeing that the value of the vertical axis at time t=4 st=4\text{ s} is +1ms2+1\dfrac{\text m}{\text s^2}.

What does the slope represent on an acceleration graph?

The slope of an acceleration graph represents a quantity called the jerk. The jerk is the rate of change of the acceleration.
For an acceleration graph, the slope can be found from slope=riserun=a2a1t2t1=ΔaΔt\text{slope}=\dfrac{\text{rise}}{\text{run}}=\dfrac{a_2-a_1}{t_2-t_1}=\dfrac{\Delta a}{\Delta t}, as can be seen in the diagram below.
This slope, which represents the rate of change of acceleration, is defined to be the jerk.
jerk=ΔaΔt\text{jerk}=\dfrac{\Delta a}{\Delta t}
As strange as the name jerk sounds, it fits well with what we would call jerky motion. If you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time, the motion would feel jerky, and you would have to keep applying different amounts of force from your muscles to stabilize your body.
To finish up this section, let's visualize the jerk with the example graph shown below. Try moving the dot horizontally to see what the slope—i.e., jerk—looks like at different points in time.
Concept check: For the acceleration graph shown above, is the jerk positive, negative, or zero at t=6 st=6\text{ s}?
The jerk is negative at t=6 st=6\text{ s} since the slope is negative. We can see this by sliding the dot to t=6 st=6\text{ s} and observing that the slope is directed downward.

What does the area represent on an acceleration graph?

The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.
area=Δv\Large \text{area}=\Delta v
It might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 ms2~\dfrac{\text m}{\text s^2} for a time of 9 s.
If we multiply both sides of the definition of acceleration, a=ΔvΔta=\dfrac{\Delta v}{\Delta t}, by the change in time, Δt\Delta t, we get Δv=aΔt\Delta v=a\Delta t.
Plugging in the acceleration 4 ms2~\dfrac{\text m}{\text s^2} and the time interval 9 s we can find the change in velocity:
Δv=aΔt=(4 ms2)(9 s)=36ms\Delta v=a\Delta t= (4~\dfrac{\text m}{\text s^2})(9\text{ s})=36\dfrac{\text m}{\text s}
Multiplying the acceleration by the time interval is equivalent to finding the area under the curve. The area under the curve is a rectangle, as seen in the diagram below.
The area can be found by multiplying height times width. The height of this rectangle is 4 ms2~\dfrac{\text m}{\text s^2}, and the width is 9 s. So, finding the area also gives you the change in velocity.
area=4 ms2×9 s=36ms\text{area}=4~\dfrac{\text m}{\text s^2} \times 9\text{ s}=36\dfrac{\text m}{\text s}
The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.
Good question. We only proved that the area under an acceleration graph for a certain time interval gives the change in velocity for that time interval for the case of constant acceleration, but the area under any acceleration graph will give the change in velocity, even if the acceleration is not constant.
The reason is that the area under an acceleration graph that is changing can be broken up into small rectangles of very thin width, as seen in the diagram below. If the width of the rectangles is small enough, the acceleration will roughly be constant. Multiplying height times width will give the area and the change in velocity, Δv=aΔt\Delta v=a\Delta t, during each small interval of time. Adding up the areas of the rectangles gives roughly the total area under the curve and the total change in velocity for a given time interval.
If the width of the rectangles is taken to be infinitesimally small, the sum of the areas of all the rectangles will give exactly the area under the curve and the exact change in velocity. So the area under any acceleration graph gives the change in velocity. If this discussion sounds like mathematical witchcraft it's probably because the proper derivation of this proof requires the ideas of calculus—the integral in calculus is an operation that allows you to find areas under curves.
Note: When solving a problem or answering a question, we don't have to actually break a curve into rectangles when finding the area. Since we know that the area under the curve is equal to the change in velocity, we are free to find the area under the curve in whichever way is most convenient. In other words, for the graph shown above, we could just use the formula for the area of a triangle, 12bh\dfrac{1}{2}bh, to find the area and change in velocity.

What do solved examples involving acceleration vs. time graphs look like?

Example 1: Race car acceleration

A confident race car driver is cruising at a constant velocity of 20 m/s. As she nears the finish line, the race car driver starts to accelerate. The graph shown below gives the acceleration of the race car as it starts to speed up. Assume the race car had a velocity of 20 m/s at time t=0 st=0\text{ s}.
What is the velocity of the race car after the 8 seconds of acceleration shown in the graph?
We can find the change in velocity by finding the area under the acceleration graph.
Δv=area=12bh=12(8 s)(6ms2)=24 m/s(Use the formula for area of triangle: )12bh.\Delta v=\text{area}=\dfrac{1}{2}bh=\dfrac{1}{2}(8\text{ s})(6\dfrac{\text m}{\text s^2})=24\text{ m/s}\quad \text{(Use the formula for area of triangle: $\dfrac{1}{2}bh.$)}
Δv=24 m/s(Calculate the change in velocity.)\Delta v=24\text{ m/s}\quad \text{(Calculate the change in velocity.)}
But this is just the change in velocity during the time interval. We need to find the final velocity. We can use the definition of the change in velocity, Δv=vfvi\Delta v=v_f-v_i, to find that
Δv=24 m/s\Delta v=24\text{ m/s}
vfvi=24 m/s(Plug in  for .)vfviΔvv_f-v_i=24\text{ m/s} \qquad{\text{(Plug in $v_f-v_i$ for $\Delta v$.)}}
vf20 m/s=24 m/s(Plug in 20 m/s for the initial velocity .)viv_f-20\text{ m/s}=24\text{ m/s} \qquad{\text{(Plug in 20 m/s for the initial velocity $v_i$.)}}
vf=24 m/s+20 m/s(Solve for .)vfv_f=24\text{ m/s}+20\text{ m/s}\qquad{\text{(Solve for $v_f$.)}}
vf=44 m/s(Calculate and celebrate!)v_f=44\text{ m/s}\qquad{\text{(Calculate and celebrate!)}}
The final velocity of the race car was 44 m/s.

Example 2: Sailboat windy ride

A sailboat is sailing in a straight line with a velocity of 10 m/s. Then at time t=0 st=0\text{ s}, a stiff wind blows causing the sailboat to accelerate as seen in the diagram below.
What is the velocity of the sailboat after the wind has blown for 9 seconds?
The area under the graph will give the change in velocity. The area of the graph can be broken into a rectangle, a triangle, and a triangle, as seen in the diagram below.
The blue rectangle between t=0 st=0\text{ s} and t=3 st=3\text{ s} is considered positive area since it is above the horizontal axis. The green triangle between t=3 st=3\text{ s} and t=7 st=7\text{ s} is also considered positive area since it is above the horizontal axis. The red triangle between t=7 st=7\text{ s} and t=9 st=9\text{ s}, however, is considered negative area since it is below the horizontal axis.
We'll add these areas together—using hwhw for the rectangle and 12bh\dfrac{1}{2}bh for the triangles—to get the total area between t=0 st=0\text{ s} and t=9 st=9\text{ s}.
Δv=area=(4ms2)(3 s)+12(4 s)(4ms2)+12(2 s)(2ms2)(Add areas of rectangle and two triangles.)\Delta v=\text{area}=(4\dfrac{\text m}{\text s^2})(3\text{ s})+\dfrac{1}{2} (4\text{ s})(4\dfrac{\text m}{\text s^2})+\dfrac{1}{2}(2\text{ s})(-2\dfrac{\text m}{\text s^2}) \quad \text{(Add areas of rectangle and two triangles.)}
Δv=18 m/s(Calculate to get total change in velocity.)\Delta v=18\text{ m/s} \quad \text{(Calculate to get total change in velocity.)}
But this is the change in velocity, so to find the final velocity, we'll use the definition of change in velocity.
vfvi=18 m/s(Use definition of change in velocity.)v_f-v_i=18\text{ m/s}\quad \text{(Use definition of change in velocity.)}
vf=18 m/s+vi(Solve for the final velocity.)v_f=18\text{ m/s}+v_i\quad \text{(Solve for the final velocity.)}
vf=18 m/s+10 m/s(Plug in initial velocity.)v_f=18\text{ m/s}+10\text{ m/s}\quad \text{(Plug in initial velocity.)}
vf=28 m/s(Calculate and celebrate!)v_f=28\text{ m/s}\quad \text{(Calculate and celebrate!)}
The final velocity of the sailboat is vf=28 m/sv_f=28\text{ m/s}.
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