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A car is travelling on a road. The maximum velocity the car can attain is $24 {ms}^{- 1}$ and 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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$6 {ms}^{- 2}$

$1.2 {ms}^{- 2}$

$12 {ms}^{- 2}$

$3.6 {ms}^{- 2}$

answer is B.

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Detailed Solution

Area of $v - t$ graph $=$ displacement
$∴ 1032 = \frac{1}{2} (56 + t_{0}) \times 24$
or $t_{0} = 30 s$

A car is travelling on a road. The maximum velocity the car can attain is 24 ms^-1 and the maximum deceleration is 4 ms^-2. If car starts from rest and comes to

Deceleration time $t_{1} = \frac{24}{4} = 6 s$
$∴$ Acceleration time $t_{2} = 56 - t_{0} - t_{1} = 20 s$
$∴$ Acceleration $= \frac{24}{20} = 1.2 m / s^{2}$

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