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1st Equation of Motion

A car is trav...

Question

A car is travelling on a straight road. The maximum velocity the car can attains is $24 m s^{- 1} .$ The maximum acceleration and deceleration it can attain are $1 m s^{- 2}$ and $4 m s^{- 2}$ respectively. The shortest time the car takes from rest to rest in a distance of 500 m is,

A

22.4 s

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B

30 s

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C

11.2 s

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D

35.8 s

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Solution

The correct option is D 35.8 s
$D i s p l a c e m e n t i f t h e c a r r e a c h e s m a x s p e e d w i t h m a x a c c e l e r a t i o n s_{1} = \frac{v^{2}}{2 a} = \frac{24^{2}}{2} = 288 m D i s p l a c e m e n t w h e n t h e c a r r e a c h e s f r o m m a x s p e e d t o z e r o s_{2} = \frac{u^{2}}{2 a} = \frac{24^{2}}{2 \times 4} = 72 s_{1} + s_{2} = 288 + 72 = 360 m r e m a i n i n g d i s t a n c e i s t r a v e l l e d a t m a x i m u m s p e e d, f o r w h i c h t h e t i m e t a k e n i s \frac{500 - 360}{24} = 5.833 s t i m e d u r i n g s_{1} = \frac{24}{1} = 24 s t i m e d u r i n g s_{2} = \frac{24}{4} = 6 s t o t a l t i m e i s 35.833 s$

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