JAMB - Physics (2025 - No. 96)

From the above graph, what is the distance covered in the last stage of the motion?

56m
28m
140m
46m

Explanation

The velocity-time graph has three stages:
\(\item \(0\text{--}10 \, \text{s}\): constant \(12 \, \text{m/s}\)
\( \item \(10\text{--}14 \, \text{s}\): deceleration from \(12 \, \text{m/s}\) to \(0 \, \text{m/s}\) (last stage)

Distance in the last stage is given by the area of the triangle:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (14 - 10) \, \text{s} \times 12 \, \text{m/s}\)
\(= \frac{1}{2} \times 4 \times 12 = 28 \, \text{m}\).

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