A train accelerates from rest at a constant r | vedclass

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(B) For the acceleration phase starting from rest: $v_{\max} = \alpha t_1$ and $x_1 = \frac{1}{2} \alpha t_1^2$. For the retardation phase coming to r

Vedclass · Accepted Answer

$\frac{x_1}{x_2} = \frac{\beta}{\alpha} = \frac{t_1}{t_2}$

Vedclass · Answer

(B) For the acceleration phase starting from rest: $v_{\max} = \alpha t_1$ and $x_1 = \frac{1}{2} \alpha t_1^2$. For the retardation phase coming to rest: $v_{\max} = \beta t_2$ and $x_2 = \frac{1}{2} \beta t_2^2$. Equating the maximum velocities: $\alpha t_1 = \beta t_2$,which implies $\frac{t_1}{t_2} = \frac{\beta}{\alpha}$. Now,taking the ratio of distances: $\frac{x_1}{x_2} = \frac{\frac{1}{2} \alpha t_1^2}{\frac{1}{2} \beta t_2^2} = \frac{\alpha}{\beta} \left( \frac{t_1}{t_2} \right)^2$. Substituting $\frac{\alpha}{\beta} = \frac{t_2}{t_1}$ into the distance ratio: $\frac{x_1}{x_2} = \left( \frac{t_2}{t_1} \right) \left( \frac{t_1}{t_2} \right)^2 = \frac{t_1}{t_2}$. Since $\frac{t_1}{t_2} = \frac{\beta}{\alpha}$,we get $\frac{x_1}{x_2} = \frac{t_1}{t_2} = \frac{\beta}{\alpha}$.

$A$ train accelerates from rest at a constant rate $\alpha$ for distance $x_1$ and time $t_1$. After that,it retards to rest at a constant rate $\beta$ for distance $x_2$ and time $t_2$. Which of the following relations is correct?

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