Question

A television camera at ground level is filming at the lift-off of a space shuttle that is rising vertically according to the position equation S = 50t 2, where S is measured in feet and t is measured in seconds. The camera is 2000 ft from the launch pad. What is the rate of change in the angle of elevation of the camera 10 seconds after lift-off?

Answer 1

Answer:

Explanation:

s = 50 t^2

Differentiate both sides with respect to t

ds/dt = 50 x 2 t

ds/dt = 100 t

Substitute t = 10 second

ds/dt = 100 x 10

ds/dt = 1000 m/s

thus, the rate of change of elevation is 1000 m/s.

Answer 2

Answer: 2/29 rad/sec

Explanation: from the attachment ,

S=50t^2

S=50×10^2

S=5000

Tan¤= opp/Adj

=5000/2000

Tan-1 2.5=68.19°

Let y be the angle of elevation

Let S be the camera height

ds/dt= 50×2t= 100t

Substituting t=10

100(10)=1000

The rate of change of angle of elevation of the camera will be calculated using :

dy/dt=1/2000÷1+(5000/2000)2 ×1000

dy/dt=1/2÷(1+25/4)

dy/dt=2/29rad/sec