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Acceleration in 2D

A train accel...

Question

A train accelerates from rest at a constant rate $α$ for distance $x_{1}$ and time $t_{1}$ . After that it retards to rest at constant rate $β$ for distance $x_{2}$ and time $t_{2}$ . Which of the following relations is correct?

\frac{x_{1}}{x_{2}} = \frac{α}{β} = \frac{t_{1}}{t_{2}}

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\frac{x_{1}}{x_{2}} = \frac{β}{α} = \frac{t_{1}}{t_{2}}

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\frac{x_{1}}{x_{2}} = \frac{α}{β} = \frac{t_{2}}{t_{1}}

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\frac{x_{1}}{x_{2}} = \frac{β}{α} = \frac{t_{2}}{t_{1}}

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Solution

The correct option is B $\frac{x_{1}}{x_{2}} = \frac{β}{α} = \frac{t_{1}}{t_{2}}$
According to question.................
$\begin{matrix} T h e l a w o f k i n e m a t i c s f o r c o n s t a n t a c c e l e r a t i o n a n d r e t a r d a t i o n, t o f i n d t h e r e l a t i o n s h i p . F o r t h e 1 s t h a l f o f a c c e l e r a t e d m o t i o n, w e c a n w r i t e, v^{2} = 2 α x_{1} . . . . . . . . . . . . . . . . . . . . . . (i) (v i s t h e v e l o c i t y g a i n e d a f t e r t i m e t_{a}) N o w, a t t h e e n d o f f i r s t t_{1} s, i f i t g a i n s v e l o c i t y v, t h e n, v = 0 + α t_{1} S o, p u t t i n g t h e v a l u e o f v i n e q u a t i o n (i) w e g e t, α^{2} (t_{1})^{2} = 2 α x_{1} o r, x_{1} = α \frac{{(t_{1})}^{2}}{2} S o, f o r t h e r e s t h a l f o f t h e j o u r n e y, w e c a n s a y, 0^{2} = (α t_{1})^{2} - 2 β x_{2} o r, x_{2} = α^{2} \frac{{(t_{1})}^{2}}{2 β} A n d, 0 = α t_{1} - β t_{2} S o, \frac{α}{β} = \frac{t_{2}}{t_{1}} A n d, \frac{x_{1}}{x_{2}} = \frac{\frac{α {(t_{1})}^{2}}{2}}{α^{2} \frac{{(t_{2})}^{2}}{2 β}} = \frac{β}{α} S o, w e c a n c o n c l u d e, \frac{x_{1}}{x_{2}} = \frac{β}{α} = \frac{t_{1}}{t_{2}} \end{matrix}$
so the option is B.

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