a particle moves in a straight line. its started from rest. it covered distance X with uniform acceleration to achieve velocity V then 2X with uniform velocity and 3X with uniform retardation . the average velocity of the particle is??

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Solution

$I n i t i a l v e l o c i t y i s z e r o, i . e . u = 0 t i m e i n w h i c h v e o c i t y b e c o m e s v i s t_{1} v = u + a t_{1} v = a t_{1} a = \frac{v}{t_{1}} d i s \tan c e t r a v e l l e d i s t_{1} s = u t + \frac{1}{2} a {t_{1}}^{2} x = 0 + \frac{1}{2} \times \frac{v}{t_{1}} {t_{1}}^{2} t_{1} = \frac{2 x}{v} w h e n b o d y m o v e s w i t h u n i f o r m v e l o c i t y v d u r i n g t i m r t_{2}, d i s \tan c e t r a v e l l e d i s 2 x v = \frac{2 x}{t_{2}} t_{2} = \frac{2 x}{v} w h e n b o d y m o v e s w i t h u n i f r o m r e t a r d a t i o n d, t i m e t a k e n i s t_{3} d i s \tan c e t r a v e l e d i s 3 x t h e n f i n a l v e l o c i t y i s 0 0 = v - d t_{3} d = \frac{v}{t_{3}} d i s \tan c e t a r v e l l e d 3 x = v t_{3} - \frac{1}{2} d {t_{3}}^{2} 3 x = v t_{3} - \frac{1}{2} \times \frac{v}{t_{3}} {t_{3}}^{2} 3 x = \frac{1}{2} v t_{3} t_{3} = \frac{6 x}{v} a v e r a g e v e l o c i t y = \frac{t o t a l d i s \tan c e t r a v e l l e d}{t o t a l t m e t a k e n} = \frac{x + 2 x + 3 x}{t_{1} + t_{2} + t_{3}} = \frac{6 x}{\frac{2 x}{v} + \frac{2 x}{v} + \frac{6 x}{v}} = \frac{6 x}{10 x} v = \frac{3}{5} v$

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