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Q.

Two objects P and Q, travelling in the same direction start from rest. While the object P starts at time $t = 0$ and object Q starts later at $t = 30 \min$ . Object P has an acceleration of $40 k m / h^{2}$ . To catch P at a distance of 20 km, the acceleration of Q should be

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a

$100 k m / h r^{2}$

b

$160 k m / h r^{2}$

c

$80 k m / h r^{2}$

d

$40 k m / h r^{2}$

answer is D.

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Detailed Solution

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P starts $t = 0 Q_{p} = 40 k m / h r^{2}$
Time taken by P to cover 20 km,
$S = u t + \frac{1}{2} a t^{2} \Rightarrow t = \sqrt{\frac{2 S}{a}} = \sqrt{\frac{2 \times 20}{40}} = 1 h o u r$
Time taken by $Q = 1 - \frac{1}{2} h o u r = 30 \min = \frac{1}{2} h r$

$∴ a_{Q} = \frac{2 s}{t^{2}} = \frac{2 \times 20}{{(\frac{1}{2})}^{2}} = 160 k m / h r^{2}$

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