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Average Acceleration — Definition, Formula & Examples

Average acceleration is how much an object's velocity changes per unit of time over a whole time interval. It tells you the overall rate of speeding up or slowing down, without describing what happens at each instant.

If v(t)v(t) is the velocity of an object, the average acceleration on [t1,t2][t_1, t_2] with t1<t2t_1 < t_2 is v(t2)−v(t1)t2−t1\dfrac{v(t_2) - v(t_1)}{t_2 - t_1}. This is the slope of the secant line through (t1,v(t1))(t_1, v(t_1)) and (t2,v(t2))(t_2, v(t_2)) on the velocity graph. Instantaneous acceleration a(t)=v′(t)a(t) = v'(t) is the limit of this quotient as t2→t1t_2 \to t_1. When aa is continuous, the average acceleration also equals the average value of a(t)a(t) on the interval, 1t2−t1∫t1t2a(t) dt\frac{1}{t_2 - t_1}\int_{t_1}^{t_2} a(t)\,dt.

Key Formula

aavg=ΔvΔt=v(t2)−v(t1)t2−t1a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}
Where:
  • aavga_{\text{avg}} = average acceleration (for example, in m/s²)
  • v(t1),v(t2)v(t_1), v(t_2) = velocity at the start and end of the interval
  • t1,t2t_1, t_2 = start and end times, with t_1 < t_2
  • Δv,Δt\Delta v, \Delta t = change in velocity and change in time

Worked Example

Problem: A particle's position in meters is s(t)=t3+2ts(t) = t^3 + 2t, where tt is in seconds. Find its average acceleration from t=1t = 1 to t=3t = 3.
Step 1: Acceleration is a rate of change of velocity, so first find velocity by differentiating position.
v(t)=s′(t)=3t2+2v(t) = s'(t) = 3t^2 + 2
Step 2: Evaluate the velocity at each endpoint.
v(1)=3(1)2+2=5,v(3)=3(3)2+2=29v(1) = 3(1)^2 + 2 = 5, \qquad v(3) = 3(3)^2 + 2 = 29
Step 3: Divide the change in velocity by the change in time.
aavg=29−53−1=242=12a_{\text{avg}} = \frac{29 - 5}{3 - 1} = \frac{24}{2} = 12
Answer: The average acceleration is 1212 m/s² on [1,3][1, 3].

Why It Matters

AP Calculus AB and BC free-response questions often give velocity in a table and ask you to estimate acceleration. The expected method is average acceleration between two table values, and you must include units such as m/s². The same secant-slope idea motivates the definition of the derivative as a limit.

Common Mistakes

Mistake: Using position values instead of velocity values. In the example above, s(3)−s(1)3−1\frac{s(3) - s(1)}{3 - 1} gives 33−32=15\frac{33 - 3}{2} = 15.
Correction: That quotient is average velocity, measured in m/s. Average acceleration needs a change in velocity. If you are given position, differentiate first to get v(t)v(t). Then use the endpoint velocities.

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