The speed of a train increases at a constant rate $α$ from zero to $v$ and then remains constant for an interval and finally decreases to zero at a constant rate $β$ . The total distance travelled by the train is $l$ . The time taken to complete the journey is $t$ . Then,

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$t = \frac{l (α + β)}{α β}$

$t = \frac{l}{v} + \frac{v}{2} (\frac{1}{α} + \frac{1}{β})$

$t$ is minimum when $v = \sqrt{\frac{2 l α β}{(α + β)}}$

$t$ is minimum when $v = \sqrt{\frac{2 l α β}{(α - β)}}$

answer is B, D.

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Detailed Solution

$ν = α t_{1} \Rightarrow t_{1} = \frac{ν}{α}$
$v = β t_{2} \Rightarrow t_{2} = \frac{v}{β}$

The speed of a train increases at a constant rate alpha from zero to v and then remains constant for an interval and finally decreases to zero at a constant rate beta.

$∴ t_{0} = t - t_{1} - t_{2} = (t - \frac{v}{α} - \frac{v}{β})$

Now, $l = \frac{1}{2} α t_{1}^{2} + v t_{0} + \frac{1}{2} β t_{2}^{2}$
$= \frac{1}{2} (α) {(\frac{v}{α})}^{2} + v (t - \frac{v}{α} - \frac{v}{β}) + \frac{1}{2} (β) {(\frac{v}{β})}^{2}$
$= v t - \frac{v^{2}}{2 α} - \frac{v^{2}}{2 β}$
$∴ t = \frac{l}{v} + \frac{v}{2} (\frac{1}{α} + \frac{1}{β})$
For $t$ to be minimum its first derivation with respect to velocity be zero, i.e.
$0 = - \frac{1}{v^{2}} + \frac{α + β}{2 α β}$
$∴ v = \sqrt{\frac{2 l α}{α + β}}$