From the above graph, what is the distance covered in the last stage of the motion?
The velocity-time graph has three stages:
0 - 10s: constant \(12 \, \text{m/s}\)
10 - 14s: deceleration from \(12 \, \text{m/s}\) to \(0 \, \text{m/s}\) (last stage)
The area of the triangle gives the distance in the last stage:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (14 - 10) \, \text{s} \times 12 \, \text{m/s}\)
\(= \frac{1}{2} \times 4 \times 12 = 24 \, \text{m}\).
There is an explanation video available below.
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