Question

A 4-foot tall child walks directly away from a 12-foot tall lamppost at 2 mph. How quickly is the length of her shadow increasing when she is 6 feet away from the lamppost (rounded to the nearest tenth of a foot per second)

Answer 1

The length of the shadow is increasing with the rate of 1.5 feet per sec

Explanation:

Let AB and CD represents the height of the lamppost and child respectively ( shown below )

Also, let E be a point represents the position of child.

In triangles ABE and CDE,

`\angle ABE\cong \angle CDE` ( right angles )

`\angle AEB\cong \angle CED` ( common angles )

By AA similarity postulate,

`\triangle ABE\sim \triangle CDE`

∵ Corresponding sides of similar triangles are in same proportion,

`\implies (AB)/(CD)=(BE)/(DE)`

We have, AB = 12 ft, CD = 4 ft, BE = BD + DE = 6 + DE,

`\implies (12)/(4)=(6+DE)/(DE)`

`12DE = 24 + 4DE`

`8DE = 24`

`DE=3`

Now, the speed of walking = 2 mph =
`(2* 5280)/(3600)\approx 2.933\text{ ft per sec}`

Note: 1 mile = 5280 ft, 1 hour = 3600 sec

Thus, the time taken by child to reach at E

`= \frac{\text{Walked distance}}{\text{Walking speed}}`

`=(6)/(2.933)`

= 2.045 hours

Hence, the change rate in the length of shadow

`= \frac{\text{Length of shadow}}{\text{Time taken}}`

`=(3)/(2.045)`

= 1.5 ft per sec.