The position function of an object moving along a straight line is given by s=f(t). The average velocity of an object over the time interval [a,b] is the average rate of change of f over [a,b]; it’s velocity at t=a is the rate of change of f at a.A ball is thrown straight up with an initial velocity of 144ft/sec, so that it’s height (in feet) after t sec is given by s=f(t)= 144t-16t^2.(A)What is the average velocity of the ball over the following time intervals?[3,4]= ft/sec[3,3.5]= ft/sec[3,3.1]= ft/sec(B) what is the instantaneous velocity at time t=3? Ft/sec(C) what is the instantaneous velocity at time t=8? Ft/sec(D) when will the ball hit the ground?

A) Position at time t = 3 s : 
s₁ = 144×3 - 16×3² = 288 ft 
Position at time t = 4 s : 
s₂ = 144×4 - 16×4² = 320 ft 

Δs = s₂ - s₁ = 320 - 288 =  32ft 
Δt = t₂ - t₁ = 4 - 3 = 1 s 

Vavg[3, 4] = Δs / Δt = 32 / 1 = 32 ft/s
Vavg[3 , 3.5] = [(144×3.5 - 16×3.5²) - (144×3 - 16×3²)] / [3.5 - 3] = 40 ft/s
Vavg[4 , 4.1] = [(144×3.1 - 16×3.1²) - (144×3 - 16×3²)] / [3.1 - 3] = 4.64 ft/s

B) lim [(144t - 16t²) - (144×4 - 16×3²)) / (t - 3)] = 
t→3
lim [(-16t² + 144t - 288) / (t - 3)] = 
t→3

48 ft/s

C) lim [(144t - 16t²) - (144×8 - 16×8²)) / (t - 8)] = 
t→8 

lim [(-16t² + 144t - 128) / (t - 8)] = 
t→8 

lim (-16t + 16) = 
t→8 

-112 ft/s 

D) 144t - 16t² = 0 
(144 - 16t) t = 0 
t = 0 
or 
144 - 16t = 0 

t = 9