An object drops from a cliff that is 150 m high. The distance, d, in metres, that the object has dropped at t seconds is modelled by d(t) 4.9t^2.

Question

(a) Find the average rate of change of distance with respect to time from 2 s to 5 s.

(b) Find the instantaneous rate of change of distance with respect to time at 4 s. 

(c) Find the rate at which the object hits the ground to the nearest tenth.

Answer 1

To figure length of a fall use: (t^2+t)/2*9.81 (meters)

32.2 (ft).

Answer 2

(a). Average rate of change = [d(5) - d(2)] / (5 - 2) = 34.3 meters/sec

(b). Instantaneous rate of change at time t = d'(t) = 9.8t. So, d'(4) = 9.8(4) = 39.2 meters/sec

(c). The object hits the ground when d(t) = 150.

4.9t² = 150 So, t = 5.53 sec

rate at which the object hits the ground = d'(5.53) = 54.2 meters/sec