Question

A man walks along a straight path at a speed of 4 ft./s. A Searchlight is located on the ground 20 feet from the path and is kept focused on the man. at what rate is the Searchlight rotating when the man is 15 feet from the point on the path closest to the searchlight?

Answer 1

Answer:

the rotation rate would be 7.27° per second

Step-by-step explanation:

we can imagine a triangle formed by the distance walked by the man along the path (L=horizontal side) the vertical distance (H=distance from the path = constant = 20 feet) and the hypotenuse as the searchlight pointing at the man with an Ф angle from the vertical side (rotation angle from the closest point)

when the man is at L1 = 15 feet from the closest point

tg Ф1 = L1/H = → Ф1 = tg ⁻¹ ( L1/H) = tg ⁻¹ ( 15 feet/20 feet ) = 36.869 °

if the man walks at v = 4 feet/s ,  0.1 seconds after he would be at

L2= L1 + v* t = 15 feet + 4 feet/s * 0.1 seconds = 15.4 feet from the closest point

then the angle of the searchlight would be

tg Ф2 = L2/H → Ф1 = tg ⁻¹ ( L2/H) = tg ⁻¹ ( 15.4 feet/20 feet ) = 37.596°

therefore when the man covers a distance from 15 feet to 15.4 feet  in 0.1 seconds , the rotation angle will vary from 36.869 ° to 37.596° . thus the rotation rate would be:

rotation rate = change of angle / time required for that change = (37.596°-36.869°)/ 0.1 seconds = 7.27° per second