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A car is travelling on a road. The maximum velocity the car can attain is ${24 ms}^{- 1}$ . The maximum deceleration is ${4 ms}^{-2}$ . If car starts from rest and comes to rest after travelling $1032 m$ in the shortest time of $56 s,$ the maximum acceleration that the car can attain is

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${12 ms}^{-2}$

${3.6 ms}^{-2}$

${6 ms}^{-2}$

${1.2 ms}^{-2}$

answer is B.

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Detailed Solution

Acceleration = 1.2 m/s2 Explanation: Given : Max velocity of - 1

Area of v-t graph = displacement

1032 = 1/2 ( 56 + t_o)(24)

t_o = 30 seconds.

Deceleration time t₁ = 24/4 = 6 seconds.

Acceleration time t₂ = 56 - t_o - t₁ = 56 - 30 - 6 = 20 seconds.

Acceleration = 24 / 20 = 1.2 m/s²

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