$A$ 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?

(A) First,convert the initial speeds from $km\, h^{-1}$ to $m\, s^{-1}$:
For the first car: $u_1 = 52 \times (1000 / 3600) \, m\, s^{-1} = 14.44 \, m\, s^{-1}$.
For the second car: $u_2 = 3 \times (1000 / 3600) \, m\, s^{-1} = 0.83 \, m\, s^{-1}$.
The distance travelled by a car is equal to the area under the speed-time graph.
Distance travelled by the first car = Area of triangle with base $5\, s$ and height $14.44 \, m\, s^{-1}$.
$= (1 / 2) \times 5 \times 14.44 = 36.1 \, m$.
Distance travelled by the second car = Area of triangle with base $10\, s$ and height $0.83 \, m\, s^{-1}$.
$= (1 / 2) \times 10 \times 0.83 = 4.15 \, m$.
Comparing the two distances,the first car travelled farther after the brakes were applied.