Question

A train accelerates from rest at a constant rate $\alpha$ for a distance ${{x}_{1}}$ and time ${{t}_{1}}$ . After that it retards to rest at a constant rate $\beta$ for distance ${{x}_{2}}$ and time ${{t}_{2}}$ . Which of the following relations is correct?
(A). $\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\alpha }{\beta }=\dfrac{{{t}_{1}}}{{{t}_{2}}}$
(B). $\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\beta }{\alpha }=\dfrac{{{t}_{1}}}{{{t}_{2}}}$
(C). $\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\alpha }{\beta }=\dfrac{{{t}_{2}}}{{{t}_{1}}}$
(D). $\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\beta }{\alpha }=\dfrac{{{t}_{2}}}{{{t}_{1}}}$

Answer 1

According to Newton’s second law of motion, force is required to change the rest of rest or motion of a body. As acceleration is constant, we can use equations of motion by substituting the corresponding values. This will give us equations in terms of $\alpha ,\,\beta ,\,{{x}_{1}},\,{{x}_{2}}$ and $\alpha ,\,\beta ,\,{{t}_{1}},\,{{t}_{2}}$ . Using these equations we can find the required relations between the given variables.

Formula used:
$v=u+at$
${{v}^{2}}={{u}^{2}}+2as$

Complete step by step solution:
When the train starts from rest, it accelerates with a constant acceleration; therefore we can use equations of motion for motion in a straight line. The equations of motion are-
$v=u+at$ -------- (1)
${{v}^{2}}={{u}^{2}}+2as$ --------- (2)
$x=ut+\dfrac{1}{2}a{{t}^{2}}$ --------- (3)
Here, $u$ is initial velocity
$v$ is final velocity
$s$ is distance travelled
$t$ is time taken
$a$ is acceleration
Here, we will consider the first case when train starts accelerating
Second case when train starts decelerating
For the first case, substituting given values in eq (1), we get,
$v=0+\alpha {{t}_{1}}$
$v=\alpha {{t}_{1}}$
The final velocity in the first case will be the initial velocity in the second case, therefore,
${{u}_{2}}=\alpha {{t}_{1}}\,,\,{{v}_{2}}=0$ --------- (3)
Substituting values in eq (1) for the second case, we get,

& v=u+at \\\ & 0=\alpha {{t}_{1}}-\beta {{t}_{2}} \\\ & \alpha {{t}_{1}}=\beta {{t}_{2}} \\\ \end{aligned}$$ $$\dfrac{\beta }{\alpha }=\dfrac{{{t}_{1}}}{{{t}_{2}}}$$ -------- (4) The relation between $$\alpha ,\,\beta ,\,{{t}_{1}},\,{{t}_{2}}$$ is $$\dfrac{\beta }{\alpha }=\dfrac{{{t}_{1}}}{{{t}_{2}}}$$ . Using eq (2) in first case we get, $$\begin{aligned} & {{v}^{2}}=0+2\alpha {{x}_{1}} \\\ & {{v}^{2}}=2\alpha {{x}_{1}} \\\ & \\\ \end{aligned}$$ -------- (5) The final velocity in first case is the initial velocity in the second case, Therefore, $${{u}_{2}}^{2}=2\alpha {{x}_{1}}\,,\,\,{{v}_{2}}=0$$ Substituting values for second case in eq (2), we get, $$\begin{aligned} & 0=2\alpha {{x}_{1}}-2\beta {{x}_{2}} \\\ & \alpha {{x}_{1}}=\beta {{x}_{2}} \\\ \end{aligned}$$ $$\dfrac{\beta }{\alpha }=\dfrac{{{x}_{1}}}{{{x}_{2}}}$$ ------ (6) From eq (4) and eq (6), we get, $$\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\beta }{\alpha }=\dfrac{{{t}_{1}}}{{{t}_{2}}}$$ Therefore the relation is $$\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\beta }{\alpha }=\dfrac{{{t}_{1}}}{{{t}_{2}}}$$ **So, the correct answer is “Option B”.** **Note:** The equations of motion can only be used when acceleration is constant; this means that the external forces acting on the system must be zero. The deceleration of a body is always negative. When starting from rest, the initial velocity is taken as zero. Similarly, when coming to rest, the final velocity is taken as zero.