Motion Graphs and Kinematic Equations

$\text{Distance}$ and $\text{speed}$ are scalars — size only. $\text{Displacement}$ and $\text{velocity}$ are the vector versions, carrying a direction as well. Acceleration is always a vector.

The distinction is not pedantry. An object thrown straight up and caught again has travelled a real distance but has zero displacement, and its average velocity over the whole flight is therefore zero while its average speed is not. Edexcel questions exploit exactly that difference.

Acceleration is the rate of change of velocity, so a change in $\text{direction}$ counts as acceleration even at constant speed — which is why an object moving in a circle is accelerating.

On a $\text{displacement–time}$ graph the gradient is the velocity. A straight line means constant velocity; a curve means the velocity is changing; a horizontal line means stationary. A negative gradient means motion back towards the origin.

On a $\text{velocity–time}$ graph the gradient is the acceleration and the area between the line and the time axis is the displacement. Area below the axis counts as negative displacement, which is how the graph records a return journey.

For a curved velocity–time graph, the acceleration at an instant is the gradient of the tangent at that point, and the displacement is found by estimating the area — counting squares is an accepted method.

When acceleration is $\text{constant}$, the graph relationships become four algebraic equations. Each omits exactly one of the five quantities, so the one to choose is the one missing the quantity you neither know nor want.

They apply only for uniform acceleration. If the acceleration changes, you must either split the motion into stages where it is constant, or work from the graph.

An object moving freely under gravity, with air resistance neglected, has constant acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ downwards. That makes it a suvat problem — but only if the signs are handled consistently.

Choose a positive direction at the start and stick to it. Taking upwards as positive makes $a=-9.81$, an upward launch velocity positive, and a downward final velocity negative. The equations then handle the whole flight, including the turning point, without needing to be restarted.

At the highest point the velocity is momentarily zero, but the acceleration is still $9.81\mathrm{m}{\mathrm{s}}^{-2}$ downwards — gravity does not switch off just because the object has stopped rising.