$A$ train starting from rest picks up a speed of $10\, m s^{-1}$ in $100\, s$. It continues to move at the same speed for the next $250\, s$. It is then brought to rest in the next $50\, s$. Plot a speed-time graph for the entire motion of the train and calculate:
$(i)$ acceleration of the train while accelerating,
$(ii)$ retardation of the train while retarding,
$(iii)$ the total distance covered by the train.

(N/A) The speed-time graph is shown in the figure.
$(i)$ Acceleration is equal to the slope of the graph during the interval $0-100\, s$:
$a = \frac{\text{Change in speed}}{\text{Time taken}} = \frac{10\, m s^{-1} - 0\, m s^{-1}}{100\, s} = 0.1\, m s^{-2}$.
$(ii)$ Retardation is equal to the magnitude of the negative slope of the graph during the interval $350-400\, s$:
$a = \frac{0\, m s^{-1} - 10\, m s^{-1}}{50\, s} = -0.2\, m s^{-2}$.
Thus,retardation is $0.2\, m s^{-2}$.
$(iii)$ The total distance covered is equal to the area under the speed-time graph (Area of trapezium $ABCD$):
$\text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$
$\text{Area} = \frac{1}{2} \times (AD + BC) \times BF$
$\text{Area} = \frac{1}{2} \times (400\, s + 250\, s) \times 10\, m s^{-1} = \frac{1}{2} \times 650\, s \times 10\, m s^{-1} = 3250\, m$.