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# A train starting from rest is accelerated and the instantaneous acceleration is given by $\dfrac{{10}} {{v + 1}}\,m{s^{ - 2}}$ . Where $v$ is the velocity in $m{s^{ - 1}}$ . Find the distance in which the train attains a velocity of $54\,kmh{r^{ - 1}}$ .A. $123.75\,m$B. $112.5\,m$C. $11.25\,m$D. $77.25\,m$

Last updated date: 17th Jul 2024
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Hint: Here we have to differentiate the given acceleration with respect to time and velocity and then take distance as the limit to find the answer.
Acceleration is velocity divided by time. Instantaneous acceleration is the rate of change of velocity with time.

Complete step by step answer:
Given,
Velocity $v = 54\,kmh{r^{ - 1}} = 15m{s^{ - 1}}$
Acceleration $a = \dfrac{{10}} {{v + 1}}m{s^{ - 2}}$
$a = \dfrac{{10}} {{v + 1}} \\ \dfrac{{dv}} {{dt}} = \dfrac{{10}} {{v + 1}} \\ v\dfrac{{dv}} {{dx}} = \dfrac{{10}} {{v + 1}} \\ \int_0^{15} {v\left( {v + 1} \right)} dv = \int_0^x {10dx} \\ \left( {\dfrac{{{v^3}}} {3} + \dfrac{{{v^2}}} {2}} \right)_0^{15} = 10x \\ x = 123.75\,m \\$

Hence, the distance in which the train attains a velocity of $54\,kmh{r^{ - 1}}$
is $123.75\,m$.