Average Acceleration

Average acceleration is the change in an object's velocity divided by the time interval in which that change occurs, measured in meters per second squared. Only the starting and ending velocities enter it, so it says nothing about what happened in between.

AP Physics 1 EK 1.2.B.3 states it as a⃗avg=Δv⃗Δt\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}, and EK 1.2.B.1 supplies the framing that gets skipped: averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time. Two endpoints, nothing else.

Check the equation sheet before you rely on it. The AP Physics 1 Table of Information does not print this definition. What it does print is F⃗net=Δp⃗Δt=mΔv⃗Δt=ma⃗\vec{F}_{net} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}, from which Δv⃗/Δt\Delta \vec{v}/\Delta t can be read back out, plus the three constant-acceleration kinematic equations.

The numerator is a vector difference. This is where the marks go. A ball hits a wall at 6.0 m/s and rebounds at 4.0 m/s along the same line. Take rightward as positive and let the contact last 0.050 s. Then Δv=(−4.0)−(+6.0)=−10.0\Delta v = (-4.0) - (+6.0) = -10.0 m/s, not 2.0 m/s, so aavg=−10.0/0.050=−200a_{avg} = -10.0/0.050 = -200 m/s2^2. Reversal makes the change larger than either speed, so subtracting the two magnitudes gives the wrong answer even though both numbers are right.

Average is not instantaneous. EK 1.2.B.5 connects them: computing an average over a very small interval gives a value very close to the instantaneous one. Over a long interval the average tells you only the net effect, so a car that speeds up hard and then coasts can share an average acceleration with one that eased up steadily. On a velocity-time graph, the average is the slope of the straight line joining the two endpoints and the instantaneous value is the slope of the tangent.

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