(1 point) A spotlight on the ground is shining on a wall 20m20m away. If a woman 2m2m tall walks from the spotlight toward the building at a speed of 0.8m/s,0.8m/s, how fast is the length of her shadow on the building decreasing when she is 8m8m from the building

The answer to the question is;

When she is 8 m from the building fast the length of her shadow on the building is decreasing at or 0.22 m/s.

Explanation:

We have

Distance of the spotlight from the building = 20 m

Distance  of woman from the building when her speed is measured = 8 m

Height of the woman = 2 m

Actual speed of the woman = 0.8 m/s

Comparing the distance of the woman from the spotlight and the wall from the spotlight, we have when the woman is 8 m from the building she is 12 m from the spotlight    

Therefore we have

where y is the shadow cast by the woman on the building = 10/3

When the woman is x distance from the building, she is 20 - x meters from the spotlight

Therefore the above equation can be written  as

 which gives finding the derivative of both sides gives

hence we have by dividing by dt gives

However we know that

Therefore

The rate of decrease of her shadow is given by

or 0.222 m/s.

where are the vectors ?

be subject to a force andchange its motion