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2.Motion in Straight Line
hard
A train accelerates from rest at a constant rate $\alpha$ for distance $x_1$ and time $t_1$. After that it retards to rest at constant rate $\beta$ for distance $x_2$ and time $t_2$. Which of the following relations is correct?
A$\frac{x_1}{x_2}=\frac{\alpha}{\beta}=\frac{t_1}{t_2}$
B$\frac{x_1}{x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}$
C$\frac{x_1}{x_2}=\frac{\alpha}{\beta}=\frac{t_2}{t_1}$
D$\frac{x_1}{x_2}=\frac{\beta}{\alpha}=\frac{t_2}{t_1}$
Solution
(b)
$x_1=\frac{1}{2} \alpha t_1^2……..(i)$
$x_2=\frac{1}{2} \beta t_2^2……..(ii)$
Further, $\quad v_{\max }=\alpha t_1……..(iii)$
and $\quad 0=v_{\text {max }}-\beta t_2……..(iv)$
From Eqs.$(iii)$ and $(iv)$, we have
$\frac{\alpha}{\beta}=\frac{t^2}{t_1}$
Then, from Eqs.$(i)$ and $(ii)$,
$\frac{x_1}{x_2}=\frac{t_1}{t_2}=\frac{\beta}{\alpha}$
$x_1=\frac{1}{2} \alpha t_1^2……..(i)$
$x_2=\frac{1}{2} \beta t_2^2……..(ii)$
Further, $\quad v_{\max }=\alpha t_1……..(iii)$
and $\quad 0=v_{\text {max }}-\beta t_2……..(iv)$
From Eqs.$(iii)$ and $(iv)$, we have
$\frac{\alpha}{\beta}=\frac{t^2}{t_1}$
Then, from Eqs.$(i)$ and $(ii)$,
$\frac{x_1}{x_2}=\frac{t_1}{t_2}=\frac{\beta}{\alpha}$
Standard 11
Physics
