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

Answer:

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

Step-by-step 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,

$\implies \frac{AB}{CD}=\frac{BE}{DE}$

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

$\implies \frac{12}{4}=\frac{6+DE}{DE}$

$12DE = 24 + 4DE$

$8DE = 24$

$DE=3$

Now, the speed of walking = 2 mph = $\frac{2\times 5280}{3600}\approx 2.933\text{ ft per sec}$

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

Thus, the time taken by child to reach at E  

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

$=\frac{6}{2.933}$

= 2.045 hours

Hence, the change rate in the length of shadow

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

$=\frac{3}{2.045}$

= 1.5 ft per sec.