a car starting from rest move with a constant acceleration and attains a velocity of the 8m/s in 8sec. it then move with a uniform velocity and finally comes to rest in 32m under uniform retardation. The total distance covered by the car is 464m. Find the acceleration, retardation and total time taken.

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Solution

Dear Student,

$T o t a l j o u r n e y c a n b e d i v i d e d a s : (i) u n i f o r m a c c e l e r a t i o n' a' f r o m v e l o c i t y 0 t o 8 m s^{- 1} i n 8 s e c . S o a c c e l e r a t i o n, a = \frac{8 - 0}{8} = 1 m s^{- 2} d i s \tan c e c o v e r e d, d_{1} = \frac{1}{2} \times 1 \times 64 = 32 m (i i) c a r m o v e s w i t h u n i f o r m v e l o c i t y 8 m s^{- 1} f o r s a y,' t' s e c t h e n, d i s t a n c e c o v e r e d, d_{2} = 8 \times t = 8 t (i i i) u n i f o r m d e c e l e r a t i o n' a_{1}' f r o m v e l o c i t y 8 t o 0 m s^{- 1} w i t h i n d i s \tan c e, d_{3} = 32 m . S o, 0^{2} = 8^{2} - 2 a_{1} \times 32 o r d e c e l e r a t i o n, a_{1} = \frac{64}{64} = 1 m s^{- 2} a n d, t i m e t a k e n f o r u n i f o r m d e c e l e r a t i o n' a_{1}' f r o m v e l o c i t y 8 t o 0 m s^{- 1}, t' = \frac{8 - 0}{a_{1}} = \frac{8}{1} = 8 s e c . N o w, g i v e n : d_{1} + d_{2} + d_{3} = 464 32 + 8 t + 32 = 464 8 t = 400 t = 50 s e c . t o t a l t i m e o f j o u r n e y, t_{t o t a l} = 8 + t + t' = 8 + 50 + 8 = 64 s e c .$

Regards,
Manoj Singh.

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