A ball is dropped from a state of rest at time t=0. The distance traveled after t seconds is s(t)=16t^2 ft.

Question

(a) How far does the ball travel during the time interval [2,2.5] ? Δs=___ ft

(b) Compute the average velocity over [2,2.5] . Δs/Δt=___ ft/sec

(c) Compute the average velocity over time intervals [2, 2.01] , [2, 2.001] , [2, 2.0001] , [1.9999, 2] , [1.999, 2] , [1.99, 2] . Use this to estimate the object's instantaneous velocity at t=2 . V(2)= ___ ft/sec

Answer 1

s(2.5) = 16(2.5)^2 = 16(6.25) = 100

s(2) = 16(2)^2 = 16(4) = 64

s(2.5)-s(2) = 100 - 64 = 36 feet in 1/2 second from t=2 to t=2.5

36/1/2 = 72 feet per second

v(t) = s'(t) = derivative of 16t^2 = 32t

v(2) = 32(2) =64

v(2.5) = 32(2.5) = 80

(80+64)/2 = 40+32= 72 feet per second= average rate of change

v(2) = 32(2) = 64 feet per second = instantaneous rate of change at time t= 2 seconds