Question

The distance an object falls (when released from rest, under the influence of Earth's gravity, and with no air resistance) is given by dt) 16t2, where d is measured in feet and t is measured in seconds. A rock climber sits on a ledge on a vertical wall and carefully observes the time it takes for a small stone to fall from the ledge to the ground a. Compute d'(t). What units are associated with the derivative, and what does it measure? b. If it takes 7 s for a stone to fall to the ground, how high is the ledge? How fast is the stone moving when it strikes the ground (in miles per hour)?

Answer 1

a) d'(t) = 32t

The units of d'(t) are ft/s and it gives how fast or how slow the object is falling.

b) The ledge is 784 ft high

And just when the stone strikes the ground, it's moving at a speed of 152.73 miles/hour

Explanation:

d(t) = 16t²

a) d'(t) = (d/dt)(d) = (d/dt) (16t²) = 32t

Since it is a time derivative, the units are ft/s and it describes the rate of change of the distance of falling object; more plainly, it gives how fast or how slow the object is falling.

b) d(t) = 16t²

If t = 7 s, how high is the ledge?

d(7) = 16 (7²) = 784 ft

How fast is the stone moving when it strikes the ground (in miles per hour)?

d'(t) = 32t

d'(7) = 32 × 7 = 224 ft/s

In miles/hour

224 ft/s = 224 ft/s × (1 mile/5280 ft) × (3600s/1 hr) = 152.73 miles/hour