The driver of a train $A$ travelling at a speed of $54\, km\, h^{-1}$ applies brakes and retards the train uniformly. The train stops in $5\, s$. Another train $B$ is travelling on the parallel track with a speed of $36\, km\, h^{-1}$. This driver also applies the brakes and the train retards uniformly. The train $B$ stops in $10\, s$. Plot the speed-time graph for both the trains on the same paper. Also,calculate the distance travelled by each train after the brakes were applied.

(N/A) For train $A$:
Initial velocity $u_A = 54\, km\, h^{-1} = 54 \times \frac{5}{18} = 15\, m\, s^{-1}$.
Final velocity $v_A = 0\, m\, s^{-1}$.
Time taken $t_A = 5\, s$.
Distance travelled by train $A$ is the area under the velocity-time graph $AB$:
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5\, s \times 15\, m\, s^{-1} = 37.5\, m$.
For train $B$:
Initial velocity $u_B = 36\, km\, h^{-1} = 36 \times \frac{5}{18} = 10\, m\, s^{-1}$.
Final velocity $v_B = 0\, m\, s^{-1}$.
Time taken $t_B = 10\, s$.
Distance travelled by train $B$ is the area under the velocity-time graph $CD$:
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10\, s \times 10\, m\, s^{-1} = 50\, m$.