Search any question & find its solution
Question:
Answered & Verified by Expert
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,
Options:
Solution:
1529 Upvotes
Verified Answer
The correct answers are:
$\frac{x}{y}=\frac{\beta}{\alpha}$, $\frac{\beta}{\alpha}=\frac{t_{1}}{t_{2}}$
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. $\quad \beta=\frac{v_{0}}{t_{2}}$
$\therefore \quad \frac{\boldsymbol{\beta}}{\boldsymbol{\alpha}}=\frac{\boldsymbol{v}_{0} / \boldsymbol{t}_{\boldsymbol{2}}}{\mathrm{v}_{0} / t_{1}}=\frac{t_{1}}{t_{2}}$
Displacement = area of graph obtained between v-t
$$
\begin{array}{l}
x=\frac{1}{2} t_{1}, v_{0} \\
y=\frac{1}{2} t_{2}, v_{0}
\end{array}
$$
$\therefore \frac{x}{y}=\frac{t_{1}}{t_{2}}
\Rightarrow \quad \frac{x}{y}=\frac{t_{1}}{t_{2}}=\frac{\beta}{a}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.