Answer

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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 $123.75\,m$.

**Additional information:**

When the time interval exceeds zero, the instantaneous acceleration is defined as the limit of the average acceleration. It is also defined with respect to time in a similar way as the derivative of velocity.

The change of velocity over a period of time is the average acceleration. Instantaneous acceleration is a change in speed over a span of time. Constant or uniform acceleration is when the velocity is constant.

An object will always have instantaneous acceleration.

Similar to velocity being the derivative of the position function, the derivative of the velocity function is instantaneous acceleration.

When a police officer pulls us over for speeding, he/ she checks the car’s instantaneous speed. Instantaneous means only one particular moment.

We need to use the limit of the rate of shift in velocity with respect to time in order to measure the acceleration, as the change in time approaches zero. Because velocity is the rate of displacement transition, then that means that the rate of a rate is acceleration. Therefore, by taking the second derivative of the displacement function, we can also identify the acceleration.

**Note:**Here we have to be very careful while taking the limit as we have first converted the units of the velocity otherwise we shall get a wrong answer.

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