Question

A man of height 1.9 m walks directly away from a lamp of height 4.75m on a level road at 6m/s. The rate at which the length of his shadow is increasing is

Options

• 1 m/s

• 2 m/s

• 3 m/s

• 4 m/s

Answer 1

4 m/s

Explanation:

Let L be the lamp and PQ be the man and
OQ = x metre be his shadow and
let MQ =y metre.

`"dy"/"dx"` = speed of the man = 6 mis (given)

Since Δ OPQ and Δ OLM are similar,

`"OM"/"OQ" = "LM"/"PQ" => (x + y)/x = 4.75/1.9 = 5/2`

`=> y = 3/2`x

∴ `"dy"/"dx" = 3/2 * "dx"/"dt"`

`=> 6 = 3/2 * "dx"/"dt"`

`=> "dx"/"dt"` = 4m/s.