Question

A driver of a car travelling at $52$ $km \;h^{–1}$ applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

Answer 1

As given in the figure below $PR$ and $SQ$ are the Speed-time graph for given two cars with initial speeds $52$ $km \;h^{–1}$ and $3$ $km \;h^{–1}$ respectively.

Distance Travelled by first car before coming to rest =Area of $\triangle$ $OPR$
= $\bigg(\frac{1}{2}\bigg) \times OR \times OP$<>= $\bigg(\frac{1}{2}\bigg)$ $\times 5 s \times 52$ $km\;h^{-1}$<>= $\bigg(\frac{1}{2}\bigg)$ $\times 5 \times \frac{(52 \times 1000) }{ 3600)} m$
= $\bigg(\frac{1}{2}\bigg)$ $\times 5\times \bigg(\frac{130 }{ 9}\bigg) m$
= $\frac{325 }{ 9}\; m$
= $36.11 \;m$
Distance Travelled by second car before coming to rest =Area of $\triangle$ $OSQ$
= $\bigg(\frac{1}{2}\bigg)$ $\times OQ \times OS$
= $\bigg(\frac{1}{2}\bigg)$ $\times 10 s \times 3 \;km\;h^{-1}$
= $\bigg(\frac{1}{2}\bigg)$ $\times 10 \times \frac{(3 \times 1000) }{ (3600)} m$
= $\bigg(\frac{1}{2}\bigg)$ $\times 10 \times \bigg(\frac{5}{6}\bigg) m$
= $5 \times \bigg(\frac{5}{6}\bigg) m$
= $\frac{25}{6}$ $m$
= $4.16 \;m$
Therefore, the entire graph depicts the car's uniform motion.