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 5m/s. Then find the distance between A and C ( in metres) at the instant all the three points are collinear.

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answer is 15.

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Detailed Solution

The length of side $C A$ at any time t is $= 20 - 5 t$

the length of side $C B$ 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 $C A$ at $t = 1$ is $20 - 5 = 15.$

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