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AP Physics 1: Algebra Notes

1.2.4 Average Acceleration

AP Syllabus focus: ‘Average acceleration equals the change in velocity divided by the time interval during which that change occurs.’

Average acceleration connects how an object’s velocity changes to how long that change takes. In AP Physics 1 Algebra, it is used to describe motion over a chosen time interval using clear sign conventions and units.

Core Idea: Average Acceleration Over a Time Interval

What average acceleration describes

Average acceleration compares the change in velocity to the elapsed time for that change. It does not require the acceleration to be steady during the interval; it is a single value that represents the overall velocity change per unit time.

Average acceleration: The change in velocity divided by the time interval during which that change occurs.

Average acceleration is a vector quantity in one dimension, meaning it can be positive or negative depending on your chosen positive direction.

The algebraic relationship

The defining relationship is:

Pasted image

Velocity–time graph for a jet car with two labeled points (P and Q) used to compute slope. The slope of the line represents acceleration, so the average acceleration over an interval is aavg=Δv/Δta_{\text{avg}}=\Delta v/\Delta t read from the rise-over-run between the two points. Source

aavg=ΔvΔta_{\text{avg}}=\dfrac{\Delta v}{\Delta t}

aavga_{\text{avg}} = average acceleration in m/s2\text{m/s}^2

Δv=vfvi\Delta v=v_f-v_i

Δv\Delta v = change in velocity in m/s\text{m/s}

Δt=tfti\Delta t=t_f-t_i

Δt\Delta t = time interval in s\text{s}

vfv_f

vfv_f = final velocity at the end of the interval in m/s\text{m/s}

viv_i

viv_i = initial velocity at the start of the interval in m/s\text{m/s}

Use consistent units (typically meters, seconds) and keep track of the signs of velocities.

Direction, Signs, and Interpretation (1D)

Choosing a positive direction

In one dimension, direction is handled by a sign convention:

  • Pick a positive direction (for example, rightward or upward).

  • Velocities in the positive direction are positive; in the opposite direction are negative.

  • Average acceleration inherits its sign from Δv\Delta v (since Δt\Delta t is positive if time moves forward).

What the sign tells you

The sign of aavga_{\text{avg}} tells you the direction of the velocity change:

  • aavg>0a_{\text{avg}} > 0: velocity becomes more positive (could mean speeding up in + direction or slowing down in − direction).

  • aavg<0a_{\text{avg}} < 0: velocity becomes more negative (could mean slowing down in + direction or speeding up in − direction).

  • aavg=0a_{\text{avg}} = 0: no net change in velocity over the interval (velocity could be constant).

Avoid the common mistake of assuming “negative acceleration” automatically means “slowing down.”

Pasted image

A set of representative vvtt graphs showing that speeding up corresponds to the velocity moving away from zero, while slowing down corresponds to the velocity moving toward zero. The four cases clarify that the sign of acceleration depends on the slope of v(t)v(t), not on whether the object is “slowing down” in everyday language. Source

Slowing down happens when velocity and acceleration have opposite signs.

Selecting the Interval and Reading Information

The interval matters

Average acceleration depends on the chosen start and end times:

  • Over a longer interval, aavga_{\text{avg}} reflects the overall change.

  • Over a shorter interval, it can be different if the motion varies.

To compute it correctly, you must identify:

  • The initial velocity at the start of the interval

  • The final velocity at the end of the interval

  • The time interval between those two moments

Practical notes for AP problems

When velocities are given with directions (or signs), treat them as signed values:

  • 3 m/s3\ \text{m/s} east” and “3 m/s-3\ \text{m/s}” can represent the same velocity if east is defined as positive.

  • If an object reverses direction during the interval, viv_i and vfv_f may have opposite signs, making Δv\Delta v large in magnitude.

Units check:

  • Velocity: m/s\text{m/s}

  • Time: s\text{s}

  • Acceleration: m/s2\text{m/s}^2 (read as “meters per second per second”)

Common Errors to Avoid

  • Using speed instead of velocity (speed has no sign; average acceleration needs velocity).

  • Mixing directions without a consistent sign convention.

  • Forgetting that Δv=vfvi\Delta v = v_f - v_i (order matters).

  • Using milliseconds or minutes without converting to seconds.

  • Treating average acceleration as describing what happens at every moment; it only summarizes the interval’s net effect.

FAQ

Yes, but convert to $\text{m/s}$ first using $1\ \text{km/h} \approx 0.278\ \text{m/s}$.

Then apply $a_{\text{avg}}=\Delta v/\Delta t$.

Then $\Delta t=0$, and average acceleration is undefined.

You must have two distinct times.

Uncertainty in $v_i$ and $v_f$ carries into $\Delta v$.

A larger relative uncertainty in $\Delta v$ gives a larger relative uncertainty in $a_{\text{avg}}$.

Not necessarily. Different choices of start/end times can produce different $\Delta v$ and thus different $a_{\text{avg}}$.

It depends on the interval.

They are equivalent unit notations.

$\text{m},\text{s}^{-2}$ is common in scientific writing and emphasises “per second squared.”

Practice Questions

Q1 (1–3 marks) A car’s velocity changes from 4.0 m/s-4.0\ \text{m/s} to +2.0 m/s+2.0\ \text{m/s} in 3.0 s3.0\ \text{s}. Calculate the average acceleration.

  • Uses Δv=vfvi\Delta v = v_f - v_i with correct substitution (1)

  • Calculates Δv=2.0(4.0)=6.0 m/s\Delta v = 2.0 - (-4.0) = 6.0\ \text{m/s} (1)

  • aavg=Δv/Δt=6.0/3.0=2.0 m/s2a_{\text{avg}}=\Delta v/\Delta t = 6.0/3.0 = 2.0\ \text{m/s}^2 with correct unit (1)

Q2 (4–6 marks) A cyclist travels along a straight road (take forward as positive). At t=0t=0, vi=+6.0 m/sv_i=+6.0\ \text{m/s}. At t=4.0 st=4.0\ \text{s}, v=+2.0 m/sv=+2.0\ \text{m/s}. At t=7.0 st=7.0\ \text{s}, vf=1.0 m/sv_f=-1.0\ \text{m/s}.
(a) Determine the average acceleration from 00 to 4.0 s4.0\ \text{s}.
(b) Determine the average acceleration from 4.04.0 to 7.0 s7.0\ \text{s}.
(c) Determine the average acceleration from 00 to 7.0 s7.0\ \text{s}.

(a) Δv=2.06.0=4.0 m/s\Delta v = 2.0 - 6.0 = -4.0\ \text{m/s} (1), Δt=4.0 s\Delta t = 4.0\ \text{s} (1), aavg=4.0/4.0=1.0 m/s2a_{\text{avg}}=-4.0/4.0=-1.0\ \text{m/s}^2 (1)
(b) Δv=1.02.0=3.0 m/s\Delta v = -1.0 - 2.0 = -3.0\ \text{m/s} (1), Δt=3.0 s\Delta t = 3.0\ \text{s}, aavg=3.0/3.0=1.0 m/s2a_{\text{avg}}=-3.0/3.0=-1.0\ \text{m/s}^2 (1)
(c) Δv=1.06.0=7.0 m/s\Delta v = -1.0 - 6.0 = -7.0\ \text{m/s} (1), Δt=7.0 s\Delta t = 7.0\ \text{s}, aavg=7.0/7.0=1.0 m/s2a_{\text{avg}}=-7.0/7.0=-1.0\ \text{m/s}^2 (1)

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