Explanation

(d) From first equation of motion, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = u + at$
Given, $u = 0,\ a_{1} = 2\ {ms}^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 10\ s,$
$\therefore\ \ \ \ v_{1} = 2 \times 10 = 20\ {ms}^{- 1}$
In the next 30 s, the constant velocity becomes
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{2} = v_{1} + a_{2}t_{2}$
Given, $v_{1} = 20\ {ms}^{- 1},\ a_{2} = 2\ {ms}^{- 2},\ t_{2} = 30\ s$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{2} = 20 + 2 \times 30 = 80\ {ms}^{- 1}.$
When it decelerates, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{3}^{2} = u^{2} - 2a_{3}s$
Here, $v_{3} = 0\ (train\ stops),\ v_{2} = 80\ {ms}^{- 1},$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{3} = 4\ {ms}^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = (80)^{2} - 2 \times 4 \times s$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = \frac{80\ \times \ 80}{8} = 800\ m.$
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