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Q.

Two particles, one with constant velocity 50 ms^-1 and the other with uniform acceleration 10 ms^-2 , start moving simultaneously from the same place in the same direction. They will be at a distance of 125 m from each other after

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a

5 s

b

$5 (1 + \sqrt{2}) s$

c

10 s

d

$10 (\sqrt{2} + 1) s$

answer is B.

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Detailed Solution

For particle 1 moving with constant velocity, $s_{1} = 50 t$
For particle 2 moving with constant acceleration $s_{2} = \frac{1}{2} {at}^{2} = \frac{1}{2} \times 10 t^{2} = 5 t^{2}$
When $s_{2} - s_{1} = 125$
The above equations $5 t^{2} - 50 t = 125$
On solving, we get $t = (5 \pm 5 \sqrt{2}) s \Rightarrow t = 5 (1 + \sqrt{2}) s$

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