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A cat and a dog are initially at a separation of 20 metre. The dog starts chasing the cat along the line joining them and at the same instant the cat starts moving away from the dog along the line joining them with an acceleration of $2.5 m / s^{2}$ . Then the minimum value of v for which the dog will be able to catch the cat is

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20 m/s

25 m/s

15 m/s

10 m/s

answer is A.

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Detailed Solution

If the dog catches the cat at point p, then

$BP = \frac{1}{2} {at}^{2} and AP = vt$

Now, Ap = AB + BP $\Rightarrow Vt = 20 + \frac{1}{2} x 2.5 x t^{2}$

$\Rightarrow 5 t^{2} - 4 vt + 80 = 0$

$∴ (- 4 v^{2}) - 4.5.80 \geq 0 \Rightarrow v \geq 10 m / s$

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