Velocity-time graphs

The idea

On a velocity-time graph, the gradient is the acceleration and the area between the graph and the time axis is the displacement. Area below the axis counts as negative displacement.

Split the graph into triangles, rectangles and trapezia.
Area $=$ displacement; gradient $=$ acceleration.
For distance, treat every area as positive.

Watch out Signed area gives displacement; distance needs every area counted positively.

See it worked

Worked example

A car speeds up uniformly from rest to $12\text{ m s}^{-1}$ in $4\text{ s}$ , travels at $12\text{ m s}^{-1}$ for $6\text{ s}$ , then slows uniformly to rest in $2\text{ s}$ . Find the distance.

Speeding up: $\frac{1}{2} \times 4 \times 12 = 24\text{ m}$ .
Steady: $6 \times 12 = 72\text{ m}$ .
Stopping: $\frac{1}{2} \times 2 \times 12 = 12\text{ m}$ . Total $108\text{ m}$ .

Answer $108\text{ m}$

Your turn

Question 1Non-calculator3 marks

A particle's velocity rises uniformly from $4\text{ m s}^{-1}$ to $16\text{ m s}^{-1}$ in $3\text{ s}$ . Find its acceleration.

Question 2Non-calculator3 marks

A cyclist accelerates uniformly from rest to $10\text{ m s}^{-1}$ in $5\text{ s}$ , then rides at $10\text{ m s}^{-1}$ for $10\text{ s}$ . Find the total distance.

Question 3Non-calculator3 marks

A train slows uniformly from $20\text{ m s}^{-1}$ to rest in $40\text{ s}$ . How far does it travel while slowing?

In past papers

19 official questions in 19 papers since 2018. The latest:

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