Question

A train starts from rest from a station with acceleration $0.2m/{s^2}$ on a straight track and then comes to rest after attaining maximum speed on another station due to retardation $0.4m/{s^2}$ . If total time spent is half an hour, then a distance between two stations is [ Neglect the length of the train]:
A) $216Km$
B) $512Km$
C) $728Km$
D) $1296Km$

Answer 1

Hint:- Firstly, we will find the relation between maximum velocity achieved and time taken for attaining maximum velocity. Then we will find the relation between time taken to stop the train (making velocity $0$ ) after attaining maximum velocity and it’s final velocity which is $0$ . Then after solving equations, we will get maximum velocity. Then using laws of motions, we can find total distance by adding distances covered by train during acceleration and during retardation.

Complete Step by Step Explanation:
Let train accelerated for time ${t_1}$ and maximum velocity be ${v_1}$
Now, according to Newton’s first law of motion,
$v = u + at$
Where $v$ is final velocity,
$u$ is initial velocity,
$a$ is acceleration,
$t$ is the time to reach velocity v from u.
So, using above equation, we get,
${v_1} = 0 + 0.2{t_1}$ (since, initial velocity of train is zero and acceleration is $0.2m/{s^2}$ )
On solving we get,
${t_1} = \dfrac{{{v_1}}}{{0.2}}$ -----(1)
Now, let velocity becomes zero from ${v_1}$ in time ${t_2}$ due to retardation of $0.4m/{s^2}.$
So, according to Newton’s first law of motion, we get,
$0 = {v_1} + ( - 0.4){t_2}$
On solving we get,
${t_2} = \dfrac{{{v_1}}}{{0.4}}$ -----(2)
Adding equation one and two, we get,
${t_1} + {t_2} = \dfrac{{{v_1}}}{{0.2}} + \dfrac{{{v_1}}}{{0.4}}$
Now, total time is given to us as half hour which is $30 \times 60 = 1800$ seconds, so we get,
$1800 = \dfrac{{{v_1}}}{{0.2}} + \dfrac{{{v_1}}}{{0.4}}$
Solving this we get,
$1800 = 7.5{v_1}$
So we get maximum velocity as,
${v_1} = 240m{s^{ - 1}}$
Now, according to Newton’s third law of motion,
${v^2} - {u^2} = 2as$
Where $v$ is final velocity,
$u$ is initial velocity,
$a$ is acceleration and
$s$ is distance covered.
So, let the distance covered during acceleration be $s_1$
So using Newton’s third law of motion,
${v_1}^2 - {0^2} = 2 \times 0.2 \times s_1$
Putting all values, we get,
${240^2} = 2 \times 0.2 \times s_1$
So, $s_1 = \dfrac{{{{240}^2}}}{{0.4}}$

So, let the distance covered during retardation be $s_2$
So using Newton’s third law of motion,
${0^2} - {v_1}^2 = 2 \times \left( { - 0.4} \right) \times s_2$
Putting all values, we get,
${240^2} = 2 \times 0.4 \times s_2$
So, $s_2 = \dfrac{{{{240}^2}}}{{0.8}}$
Total distance covered by train is $s_1 + s_2 = \dfrac{{{{240}^2}}}{{0.4}} + \dfrac{{{{240}^2}}}{{0.8}}$
On solving we get,
$s_1 + s_2 = {240^2}\left[ {\dfrac{1}{{0.4}} + \dfrac{1}{{0.8}}} \right]$
On simplifying, we get,
$s_1 + s_2 = {240^2} \times \dfrac{3}{{0.8}}$ metres,
On further solving we get,
$s_1 + s_2 = 216000$ metres
This is equivalent to $s_1 + s_2 = 216Km.$

So the correct answer is option (A).

Note: When the acceleration becomes negative (retardation), the train will still move in forward direction. Only velocity decreases during retardation, and distance will only decrease when velocity becomes negative. Hence, the train will also move forward during retardation and finally the velocity becomes zero and it doesn’t move anymore.