2.1 - Acceleration and velocity-time graphs

Acceleration is a change in velocity

Can an object have zero velocity and still be accelerating? A ball thrown straight upwards does exactly that at its highest point. It is momentarily at rest, but its velocity is changing from upwards to downwards. In the ideal free-fall model, its acceleration remains downwards throughout.

Acceleration measures how quickly velocity changes. It is a vector, because velocity includes direction. A change in direction can therefore be an acceleration even when speed stays constant; you will explore that in circular motion at A Level.

Choose the positive direction before calculating a signed change. If the initial velocity is $u$ and the final velocity is $v$, the change is final minus initial, $v-u$.

Acceleration and velocity with the same sign mean increasing speed in straight-line motion. Opposite signs mean decreasing speed. “Negative acceleration” therefore does not automatically mean slowing down. A car moving west and speeding up has both negative velocity and negative acceleration if east is positive.

For the ball at the top, “zero velocity” describes that instant; “zero acceleration” would say that its velocity is not changing at that instant. These are different claims. Free fall without air resistance has acceleration approximately $9.81\,\mathrm{m\,s^{-2}}$ downwards near Earth's surface, whether the ball is rising, at the top or falling.

Gradients and areas of velocity-time graphs

A velocity–time graph puts velocity on the vertical axis. Its height therefore describes motion, while its gradient describes the change in that motion.

A straight section has constant acceleration, $a=\Delta v/\Delta t$. A horizontal section has zero acceleration, even if its velocity is large. A downward slope has negative acceleration, whether the graph is above or below the time axis.

The area has a different meaning. For a constant velocity, a strip of width $\Delta t$ has area $v\Delta t$, which equals displacement. Adding strips gives the displacement for changing velocity too.

[DIAGRAM: asset_name: Signed areas on a velocity graph; asset_slug: motion-velocity-areas; caption: Equal positive and negative areas can give zero displacement without zero distance; description: A velocity-time graph runs horizontally from (0,4) to (3,4), slopes to (7,-4), then runs horizontally to (10,-4). Time is in seconds and velocity in metres per second. The slope crosses zero at 5 seconds. Areas above the time axis are pale blue and areas below are pale clay. The middle two triangles each have area magnitude 4 metres.]

A speed–time graph has no negative speed. Its area gives distance. Its gradient tells you the rate of change of speed; it is not enough to determine the vector acceleration when the direction of travel is changing.

Use a fresh graph described completely by coordinates.

Changing acceleration and curved graphs

When a velocity–time graph curves, acceleration changes. A single end-to-end gradient then gives an average acceleration, which may differ from the acceleration at the instant you want.

To find instantaneous acceleration, draw a tangent at the required point. Choose two widely separated points on the tangent and divide the velocity difference by the time difference. The triangle belongs to the tangent, rather than the curve. There is no need to differentiate an equation to use this method.

[DIAGRAM: asset_name: A tangent gives instantaneous acceleration; asset_slug: motion-tangent-acceleration; caption: The curve's height is velocity; the tangent's gradient is acceleration; description: A velocity-time curve v equals 8t minus t squared is plotted from 0 to 4 seconds, with velocity rising from 0 to 16 metres per second and becoming less steep. At (3,15), a clay tangent v equals 2t plus 9 is drawn. A dashed triangle uses tangent points (1,11) and (4,17), with changes 3 seconds and 6 metres per second.]

The displacement under a curved graph can be estimated with narrow trapezia. For successive speeds $v_1$ and $v_2$, a strip of width $\Delta t$ has area approximately $\tfrac12(v_1+v_2)\Delta t$. Here each velocity is in $\mathrm{m\,s^{-1}}$, so the strip gives displacement in metres. This geometric estimate is not on the data sheet: learn the area method. Make strips narrower where the curvature is stronger; do not pretend a changing curve is one straight line.

For example, a positive velocity curve has readings $0,7,12,15,16\,\mathrm{m\,s^{-1}}$ at $0,1,2,3,4\,\mathrm{s}$. Four one-second trapezia give

This is an approximation based on the supplied samples. The actual curved graph need not pass along each straight strip edge. If the graph crosses zero, split or account for signed areas before deciding whether you want displacement or distance.

Quick Recap

• Acceleration is the rate of change of velocity, with a direction.
• The gradient of a velocity–time graph is acceleration; a tangent gives its instantaneous value on a curve.
• Signed area gives displacement; adding area magnitudes gives distance.
• Compare acceleration with velocity to decide whether speed is increasing or decreasing.

At the top of an ideal vertical throw, velocity is zero but downward acceleration is not: the velocity is still changing from upwards to downwards.