Question

A passenger is standing on the platform at the beginning of the $3$ coaches of a train. The train starts moving with constant acceleration. The third coach passes by the passenger in $\Delta {t_1} = 5.0\,\sec $ and rest of the train (including the ${3^{rd}}$ coach) in $\Delta {t_2} = 20\,\sec $. In what time interval $\Delta t$ did the last coach pass by the passenger?

Answer 1

Hint
First the length of the third coach can be determined by using the second equation of the motion by using the time given in the question and then the remaining length of the train can be determined by using the same equation. By dividing these two equations, the number of coaches can be determined. Then the equation of the motion for the remaining coaches can be written, by using this equation, the time can be determined.
The equation of the motion is given by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Where, $s$ is the distance, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration.

Complete step by step answer
Given that, The time taken by the third coach to pass the passenger is, $t = 5\,\sec $
The time taken from the third coach to the last coach to pass the passenger is, $t = 20\,\sec $
Now,
The equation of the motion is given by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}\,.................\left( 1 \right)$
By substituting the initial velocity, time and acceleration in the above equation, then the above equation is written as,
$\Rightarrow s = \left( {0 \times 5} \right) + \dfrac{1}{2}a \times {5^2}$
On simplification, then the abo equation is written as,
$\Rightarrow s = \dfrac{{25}}{2}a\,...................\left( 2 \right)$
Let assume the total number of coaches is $n$, by the given information including the third coach time taken is $20\,\sec $. That means the first two coaches are not considered, so the time $20\,\sec $ is for $n - 2$ coaches.
Now, the equation of motion is written as,
$\Rightarrow \left( {n - 2} \right)s = ut + \dfrac{1}{2}a{t^2}$
By substituting the initial velocity, time and acceleration in the above equation, then the above equation is written as,
$\Rightarrow \left( {n - 2} \right)s = \left( {0 \times 20} \right) + \dfrac{1}{2}a \times {20^2}$
On simplification, then the abo equation is written as,
$\Rightarrow \left( {n - 2} \right)s = 200a\,................\left( 3 \right)$
Now the equation (2) is divided by the equation (3), then
$\Rightarrow \dfrac{s}{{\left( {n - 2} \right)s}} = \dfrac{{\dfrac{{25}}{2}a}}{{200a}}$
On cancelling the same terms, then
$\Rightarrow \dfrac{1}{{\left( {n - 2} \right)}} = \dfrac{{\dfrac{{25}}{2}}}{{200}}$
By rearranging the terms, then
$\Rightarrow 200 = \dfrac{{25}}{2}\left( {n - 2} \right)$
On further, then the above equation is written as,
$\Rightarrow 400 = 25n - 50$
By keeping the $n$ in one side and the other terms in other side, then
$\Rightarrow \dfrac{{400 + 50}}{{25}} = n$
On adding the numerator, then
$\Rightarrow \dfrac{{450}}{{25}} = n$
On dividing the above equation, then
$\Rightarrow n = 18$
Therefore, there are $18$ coaches in the train.
Already $3$ coaches pass the passenger, then for the remaining coaches the equation of motion is,
$\Rightarrow 15s = ut + \dfrac{1}{2}a{t^2}$
The initial velocity is zero, then
$\Rightarrow 15s = \dfrac{1}{2}a{t^2}\,...................\left( 4 \right)$
Now, the equation (2) divided by the equation (4), then
$\Rightarrow \dfrac{s}{{15s}} = \dfrac{{\left( {\dfrac{{25}}{2}a} \right)}}{{\left( {\dfrac{1}{2}a{t^2}} \right)}}$
By cancelling the same terms, then
$\Rightarrow \dfrac{1}{{15}} = \dfrac{{25}}{{{t^2}}}$
By rearranging the terms, then
$\Rightarrow {t^2} = 25 \times 15$
On multiplying, then
$\Rightarrow {t^2} = 375$
By taking square root on both sides, then
$\Rightarrow t = 19.3\,\sec $
The last coach passes the passenger by $ \Rightarrow 25 - 19.3$
$\Rightarrow \Delta t = 5.6\,\sec $

Note
The equation of the given by Newton's second law of motion, this equation is applicable for the linear velocity and the uniform acceleration, the condition given in the question is matched with the conditions of the Newton’s second law of the motion equation.