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Two fixed points A and B are 20 metres apart. At time $t = 0,$ the distance between a third point C and A is 20 metres and the distance between C and B is 10 metres. The component of velocity of point C along both CA and CB at any instant is $5 m / s .$ Then the distance between A and C at the instant all the three points are collinear will be

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15 m

5 m

10 m

None of these

answer is C.

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Detailed Solution

The length of side CA at any time t is = 20 – 5 t

The length of side CB at any time t is = 10 – 5 t

At the instant A,B and C are collision

(20 – 5 t) + (10 – 5 t) = 20.

Solving we get t = 1.

Therefore length of CA at t = 1 is 20 – 5 = 15.

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