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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 nert $50\, s$. Plot a speed$-$time graph for the entire motion of the train.
$(i)$ acceleration of the train while accelerating,
$(ii)$ retardation of the train while retarding,
$(iii)$ and the total distance covered by the train.
Solution
The speed$-$time graph is as shown
$(i)$ Acceleration is equal to the slope of the graph
$AB =\frac{ BF }{ AF }=\frac{10}{100}=0.1 m s ^{-2}$
$(ii)$ Retardation is equal to the slope of the graph
$CD =\frac{ CE }{ DE }=\frac{10}{50}=0.2 m s ^{-2}$
$(iii)$ Distance covered during retardation Area of trapezium $ABCD$
$=\frac{1}{2}( AD + BC ) \times BF$
$=\frac{1}{2}(400+250) \times 10=3250 m$
