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A driver of a car travelling at $52\, km\, h^{-1}$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in $5\, s$. Another driver going at $3 \,km \,h^{-1}$ in another car applies his brakes slowly and stops in $10\, s$. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied ?
Solution

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 $\Delta OPR $
$=(1 / 2) \times OR \times OP$
$=(1 / 2) \times 5\, s \times 52\, km\,h^{ -1}$
$=(1 / 2) \times 5 \times(52 \times 1000) / 3600) \,m$
$=(1 / 2) \times 5 \times(130 / 9) \,m$
$=325 / 9 \,m$
$=36.11\, m$
Distance Travelled by second car before coming to rest $=$ Area of $\Delta OSQ $
$=(1 / 2) \times O Q \times OS$
$=(1 / 2) \times 10\, s \times 3 \,km\,h^{ -1}$
$=(1 / 2) \times 10 \times(3 \times 1000) / 3600) \,m$
$=(1 / 2) \times 10 \times(5 / 6) \,m$
$=5 \times(5 / 6)\, m$
$=25 / 6 \,m$
$=4.16\, m$