Q no. 2:

2. For a particle undergoing rectinear motion with uniform acceleration, the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude of initial velocity for this time interval. Then the ratio of initial speed to the final speed for this time interval is.

$(A) \sqrt{2} (B) 2 (C) \sqrt{3} (D) 3$

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Solution

Dear Student,

$d i s p l a c e m e n t = \frac{d i s \tan c e}{3} f o r a t i m e i n t e r v a l t l e t t h e i n i t i a l v e l o c i t y b e u a n d t h e f i n a l b e v t h e d i s \tan c e i s n o t e q u a l t o d i s p l a c e m e n t i n r e c t i l i l e a r i n o n e c a s e w h e n t h e a c c n a n d i n i t i a l v e l o c i t y a r e i n o p p o s i t e d i r e c t i o n . l e t t h e d i s p l a c e m e n t b e S a n d d i t s n a c e t r a v e l b e d S = u t - \frac{a t^{2}}{2}, v = u - a t v^{2} = u^{2} - 2 a S S = \frac{u^{2} - v^{2}}{2 a} 0 = u^{2} - 2 a d' d' = \frac{u^{2}}{2 a} d'' = \frac{v^{2}}{2 a} d = \frac{u^{2} + v^{2}}{2 a} \frac{u^{2} - v^{2}}{2 a} = \frac{u^{2} + v^{2}}{6 a} 3 u^{2} - 3 v^{2} = u^{2} + v^{2} 2 u^{2} = 4 v^{2} \frac{u}{v} = \sqrt{2} o p t i o n A R e g a r d s$

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Similar questions

Q. Statement - 1 : For a particle undergoing rectilinear motion with constant acceleration, the average velocity cannot be equal for two different time intervals.
Statement -2 : The average velocity of particle moving with constant acceleration in a time interval is

\frac{\to u + \to v}{2}

,
where

\to u

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\to v

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Explain clearly, with examples, the distinction between:
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a

, initial and final velocities in a time interval

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are

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