Question

A man of height 180 cm is moving away from a lamp post at the rate of 1.2 meters per second. If the height of the lamp post is 4.5 meters, find the rate at which
(i) his shadow is lengthening
(ii) the tip of the shadow is moving

Answer 1

Let OA be the lamp post, MN be the man, MB = x be the length of the shadow and OM = y be the distance of the man from the lamp post at time t.

Then,

`("d"y)/"dt"` = 1.2 m/sec, MN = 180 cm = 1.8 m, OA = 4.5 m        .......[Given]

(i) ∆NMB ∼ ∆AOB

∴ `"MB"/"MN" = "OB"/"OA"`

∴ `x/(1.8) = (x + y)/(4.5)`

∴ 4.5x = 1.8x + 1.8y

∴ 2.7x = 1.8y

∴ x = `(1.8y)/(2.7)`

= `(2y)/3`

Differentiating w.r.t. t, we get

`("d"x)/("dt") = 2/3 xx ("d"y)/("dt")`

= `2/3 xx 1.2`

= 0.8 m/sec

(ii) B is the tip of the shadow and it is at a distance of (x + y) from the lamp post.

`"d"/"dt"(x + y) = ("d"x)/"dt" + ("d"y)/"dt"`

∴ `"d"/"dt"(x + y)` = 0.8 + 1.2

= 2 m/sec

Thus, the shadow is lengthening at the rate of 0.8 m/sec and its tip is moving at the rate of 2 m/sec.