A man stands under a lamppost that is 3.6 m high. If the man is 1.5 m tall and walks towards the lamp at a speed of 2 m/s, at what rate is the shadow lengthening after 3 sec?

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thilina-g | College Teacher | (Level 1) Educator

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Let the distance to man from the lamp post be x and the angle that shadow makes with the lamp post is alpha.

`tan(alpha) = (3.6-1.5)/x = 2.1/x`

Let the length of the shadow be y, then,

`tan(alpha) = 3.6/(x+y)`

Therefore,

`3.6/(y+x) = 2.1/x`

`12/(y+x) = 7/x`

`12x = 7y+7x`

`7y = 5x`

`y = 5/7x`

Differentiating wrt t,

`(dy)/(dt) = 5/7 (dx)/(dt)`

If the man is walking towards the lamp post at speed of `2 ms^(-1)` , then,

`(dx)/(dt) = -2`

(This is minus since x is increasing in the away direction from the lamp post).

Therefore,

`(dy)/(dt) = -10/7`

Therefore shadow reduces.

Therefore if man is walking towards that lamp at a constant rate of `2 ms^(-1) ` , the length of shadow reduces at constant `1.428 ms^(-1)` speed.

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