WBJEE · Physics · Motion In One Dimension
A train moves from rest with acceleration 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 \(\frac{x}{y}=\frac{\beta}{\alpha}\)
- B \(\frac{\beta}{\alpha}=\frac{t_{1}}{t_{2}}\)
- C \(x=y\)
- D \(\frac{x}{y}=\frac{\beta t_{1}}{\alpha t_{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\beta}{\alpha}=\frac{t_{1}}{t_{2}}\)
Step-by-step Solution
Detailed explanation
Shape of the graph \(\mathrm{OP}\) shows the acceleration in train \(\therefore \tan \theta=\) acceleration. \[ \boldsymbol{\alpha}=\frac{V_{0}}{t_{1}} \] Shape of the graph pt shows the deceleration in train. \(\therefore\) Similarly. \(\quad \beta=\frac{v_{0}}{t_{2}}\)…
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