Motion in a straight line

What you will understand

Read the axes first, distinguish a graph value from a graph slope, and calculate distance and displacement when a particle reverses direction. This focused lesson covers one-dimensional motion; it is not a complete chapter summary.

Quick answer: choose the quantity before calculating

On a position–time graph, the ordinate is position and the tangent slope is instantaneous velocity. On a velocity–time graph, slope is acceleration and signed area is displacement. For total distance, add the magnitudes of areas after splitting at direction changes. Average velocity is $\Delta x/\Delta t$; average speed is total distance divided by elapsed time. Negative acceleration alone does not establish that speed is decreasing.

Compare graph height with graph slope

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Find the reversal before adding distances

Set $v=0$ to find the turning time: $4-2t=0$ gives $t=2\,\mathrm s$. The particle reaches $x=4\,\mathrm m$, then returns toward the origin. Before this time, velocity is positive and acceleration negative, so speed decreases. After it, both are negative, so speed increases. At the turning instant speed is zero; avoid using an interval sign rule as if the velocity were nonzero there.

Add velocity–time areas with their signs

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Common mistakes to catch

A negative velocity means motion along the chosen negative direction, not a negative speed. A graph below the time axis can give negative displacement even while the particle travels a positive distance. A stationary instant can occur during a journey whose average speed is nonzero. Convert units before multiplying graph-axis values to calculate an area.

Try it yourself 1: read the direction

For the worked motion at $t=3\,\mathrm s$, which statement is correct? A. Positive position means positive velocity. B. Position is $+3\,\mathrm m$ and velocity is $-2\,\mathrm{m\,s^{-1}}$. C. Negative velocity means negative speed.

Try it yourself 2: which average uses distance?

Over the full four-second trip, which pair is correct? A. average velocity 0 and average speed 0; B. average velocity 0 and average speed 2 m/s; C. both 4 m/s.

A useful check on your next visit

Cover the result boxes. Trace the journey from the origin to the turning point and back, then explain why the average velocity is zero while the average speed is not. Use the distinction you missed to choose your next practice question.

Choose your next chapter task

Explain your self-check mistake in one sentence, then try chapter questions using the corrected rule.

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Sources and review

Technical review: Codex (automated). Sources checked on 23 October 2026. Diagrams, worked numerical models and self-checks are original teaching material. Human subject review is not claimed.