Question

A train moves from rest with acceleration $\alpha $ and in time \[{t_1}\] covers a distance $x$. It then decelerates to rest at constant retardation $\beta $ for distance $y$ in time \[{t_2}\]. Then
(A). $\dfrac{x}{y} = \dfrac{\beta }{\alpha }$
(B). $\dfrac{\beta }{\alpha } = \dfrac{{{t_1}}}{{{t_2}}}$
(C). $x = y$
(D). $\dfrac{x}{y} = \dfrac{{\beta {t_1}}}{{\alpha {t_2}}}$

Answer 1

Hint: In this question, use the second equation of motion i.e. $s = ut + \dfrac{1}{2}a{t^2}$ . Then use the initial velocity as zero and use the respective constraints of the question to find the distance travelled by the train and then use the concept that velocity is the rate change of position. Use this to approach the problem statement.

Complete step-by-step solution -

As we know from equation of motion, $s = ut + \dfrac{1}{2}a{t^2}$ , where $s$ is the displacement of the body, $u$ is the initial velocity of motion, $t$ is the total time taken and $a$ is the acceleration of the body. When the train starts moving, it is provided in the question that the train moves from rest which means it did not have any initial velocity due to which we will take $u$=0. Acceleration is provided to be $\alpha $, displacement is given $x$ and the initial time taken here is given as ${t_1}$. Now substituting these values in the above equation of motion, we get
$ \Rightarrow $ $x = 0 + \dfrac{1}{2}\alpha {t_1}^2$
$ \Rightarrow $$x = \dfrac{1}{2}\alpha {t_1}^2$ ……(1)
We have known that the velocity is the derivative of the distance with respect to time. Therefore, differentiating this equation, we get
$ \Rightarrow $$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}(\dfrac{1}{2}\alpha {t_1}^2)$
Let the velocity acquired here is $v_1$.
$ \Rightarrow {v_1} = \dfrac{1}{2}\dfrac{d}{{dt}}\alpha {t_1}^2$
$ \Rightarrow {v_1} = \alpha {t_1}$ …..(2)
Now coming to the second part of the question, where the train decelerates, we infer that the train comes to rest which means that the final velocity of the train is zero. Before decelerating, the value of initial velocity will be taken as ${v_1}$ and the final velocity i.e. ${v_{}}$ shall be taken zero. Here the displacement is given $y$ . Also as the train decelerates, the value of acceleration will be taken negative i.e. $ - \beta $ .
From another equation of motion, i.e. \[{v^2} - {u^2} = 2as\]
Substituting the values, we get
$ \Rightarrow 0 - {v_1}^2 = - 2\beta y$
$ \Rightarrow {v_1}^2 = 2\beta y$
Substituting the value of ${v_1}$ from equation (2), in the above equation, we get
$ \Rightarrow {\left( {\alpha {t_1}} \right)^2} = 2\beta y$
$ \Rightarrow y = \dfrac{{{{\left( {\alpha {t_1}} \right)}^2}}}{{2\beta }}$ ….(3)
Now from equation, (1) and (3) we get
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\dfrac{1}{2}\alpha {t_1}^2}}{{\dfrac{{{{\left( {\alpha {t_1}} \right)}^2}}}{{2\beta }}}}$
\[ \Rightarrow \dfrac{x}{y} = \dfrac{\beta }{\alpha }\] …..(4)
Also, we know that $v = u + at$
Here, the final velocity is zero and initial velocity is ${v_1}$ , and the acceleration is $ - \beta $ , and time taken is given ${t_2}$ . Substituting these values in this equation, we get
$ \Rightarrow 0 = {v_1} + ( - \beta ){t_2}$
$ \Rightarrow {v_1} = \beta {t_2}$
Now substituting the value of ${v_1}$ from equation (2), in this equation, we get
$
   \Rightarrow \alpha {t_1} = \beta {t_2} \\
   \Rightarrow \dfrac{\beta }{\alpha } = \dfrac{{{t_1}}}{{{t_2}}} \\
$
By this result and equation (4), we infer that options (A) and (B) are correct.

Note- The trick point here was the use of equations of motions. There are in general three equations of motions i.e. $s = ut + \dfrac{1}{2}a{t^2}$, $v = u + at$ and \[{v^2} - {u^2} = 2as\]. The important point here is that equations of motion are only applicable when the acceleration during the entire journey remains constant.
Negative signs for deceleration should be kept in mind. The initial velocity for the body starting from rest is zero and final velocity for stopping at the end, shall also be zero.