Question

A train stops at two stations \[s\] distance apart and takes time \[t\] on the journey from one station to the other. Its motion is first of uniform acceleration \[a\] and then immediately of uniform retardation \[b\], then the relation between $a,b,s,t\,\,{\text{is}}\,\,\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{{k{t^2}}}{{2s}}$. The value of \['k'\] is

Answer 1

Hint: In this solution, we will calculate the distance traveled by the train when it is accelerating and when it is accelerating. The total distance traveled by train can also be calculated by the total distance traveled and the time taken.

Formula used: In this solution, we will use the following formula:
Third equation of kinematics: ${v^2} - {u^2} = 2ad$where $d$ is the distance travelled by the truck with an initial velocity $u$ , final velocity $v$ , and acceleration $a$.

Complete step by step answer:
We‘ve been given that a train stops at two stations and takes time \[t\] to travel between the two stations. We will assume that the train starts from rest. Then in time ${t_1}$ it will accelerate towards the other station and have a velocity of $v = a{t_1}$
The distance travelled in this duration will be
${d_1} = \dfrac{{{v^2}}}{{2a}}$
Then the train starts retarding with a deceleration $b$ with an initial velocity $v$, then the distance it covers in time ${t_2}$, will be
${d_2} = \dfrac{{{v^2}}}{{2b}}$ and the velocity will be ${v_2} = b{t_2}$
Since the net distance travelled by the train will be $s$, we can write $s = {d_1} + {d_2}$. Substituting the value of ${d_1} = \dfrac{{{v^2}}}{{2a}}$ and ${d_2} = \dfrac{{{v^2}}}{{2a}}$, we get
$s = \dfrac{{{v^2}}}{{2a}} + \dfrac{{{v^2}}}{{2a}}$
Which gives us
$2s = \dfrac{{{v^2}}}{a} + \dfrac{{{v^2}}}{b}$
The net distance for both cases can also be calculated as the ratio of distance and time as $v = \dfrac{{2s}}{t}$ . Substituting the value of $v$ in the above equation, we get
$2s = \dfrac{{4{s^2}}}{{{t^2}a}} + \dfrac{{4{s^2}}}{{{t^2}b}}$
The above equation can be rearranged to write:
$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{{{t^2}}}{{2d}}$
Hence, the value of $k$ as seen from the equation in the question will be $k = 1$.

Note: We must realize that the final velocity of the train at the first station will be equal to the initial velocity of the train when it leaves for the second station. Also, we have assumed here that the train does not stop at the first station and just changes the form of acceleration to retardation.