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Question

A man of height 2 metre walks at uniform speed of 3 metre per second away from the lamp post of height 5 metre. Then the rate at which the length of his shadow increases is m/sec

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Solution


Let AB be the height lamp post and MN be the height of the man.
Its shadow will be y m.
We know that,
AB=5 m, MN=2 m
Assuming,
AM=x, MC=y
Then dxdt=3 m/s

Since, ABC and MNC are similar.
MNAB=MCAC
25=yx+y
y=23x
Differentiating w.r.t. to t, we get
dydt=23×dxdt dydt=23×3=2 m/sec


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