Free Pearson Edexcel International A-level Mathematics · M1

Straight-line kinematics and motion graphs

What you’ll learn

Use constant-acceleration equations with consistent signs.
Interpret displacement, velocity and acceleration graphs.

Understand the idea

Choose a positive direction before assigning velocities, displacements and accelerations. The standard constant-acceleration equations connect $u,v,a,s,t$ ; select an equation that contains the needed quantity and the available information. They apply only while acceleration is constant.

The gradient of a displacement–time graph is velocity, and the gradient of a velocity–time graph is acceleration. Signed area under a velocity–time graph is displacement. Total distance requires the magnitudes of areas on both sides of the time axis, or splitting the motion at each change of direction.

Key formulae

$v=u+at$

$s=ut+\frac12at^2$

$v^2=u^2+2as$

$s=\frac12(u+v)t$

One acceleration can first slow motion, then reverse it

Conditions. Computed v = 6 − 2t for constant acceleration −2 m s⁻² over 0–4 s. Positive velocity is motion along the chosen positive direction.

Time / s; Signed velocity / m s⁻¹.

The gradient stays −2 m s⁻². Speed decreases until t = 3 s, when v = 0, then increases in the negative direction. Zero graph height is instantaneous rest; it does not imply zero acceleration.

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

A car moving at $12\,\mathrm{m\,s^{-1}}$ slows uniformly at $3\,\mathrm{m\,s^{-2}}$ . Find its stopping distance.

Take forward as positive: $u=12$ , $v=0$ , $a=-3$ .
Use $v^2=u^2+2as$ to give $0=144-6s$ .
Solve $6s=144$ .

Answer: $24\,\mathrm m$ .

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