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

Hint: 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,
 \[\begin{align}
  & v=u+at \\
 & 0=\alpha {{t}_{1}}-\beta {{t}_{2}} \\
 & \alpha {{t}_{1}}=\beta {{t}_{2}} \\
\end{align}\]
 \[\dfrac{\beta }{\alpha }=\dfrac{{{t}_{1}}}{{{t}_{2}}}\] -------- (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,
 \[\begin{align}
  & {{v}^{2}}=0+2\alpha {{x}_{1}} \\
 & {{v}^{2}}=2\alpha {{x}_{1}} \\
 & \\
\end{align}\] -------- (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,
 \[\begin{align}
  & 0=2\alpha {{x}_{1}}-2\beta {{x}_{2}} \\
 & \alpha {{x}_{1}}=\beta {{x}_{2}} \\
\end{align}\]
 \[\dfrac{\beta }{\alpha }=\dfrac{{{x}_{1}}}{{{x}_{2}}}\] ------ (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.