The position function of an object moving along a straight line is given by s =

Question

The position function of an object moving along a straight line is given by s = f(t). The average velocity of the object over the time interval [a, b] is the average rate of change of f over [a, b]; its (instantaneous) velocity at t = a is the rate of change of fat a. A ball is thrown straight up with an initial velocity of 128 ft/sec, so that its height (in feet) after t sec is given by s = f(t) = 128t - 16^2. (a) What is the average velocity of the ball over the following time intervals? (b) What is the instantaneous velocity at time t = 4? ft/sec (c) What is the instantaneous velocity at time t = 7? ft/sec Is the ball rising or falling at this time? rising falling (d) When will the ball hit the ground? t = sec

Explanation / Answer

(a)Given s=f(t)=128t-16t^2

Vavg = ds/dt

s(4) = 256

s(5) = 240

For time interval 5 t 4:

Vavg =(s(5)-s(4))/(5-4)=(240 - 256)/(5 - 4) = -16 ft/sec => the minus sign is the direction of velocity or down

For time interval 4.5 t 4  

Vavg =(s(4.5)-s(4))/(4.5-4)=(252 - 256)/(4.5 - 4)= -8 ft/sec

For time interval 4.1 t 4  

Vavg =(s(4.1)-s(4))/(4.1-4)=(255.84 - 256)/(4.1 - 4)= -1.6 ft/sec

(b) Instanteneous velocity=V(t) = ds/dt = 128 - 32t

V(4) = 128 - 128 = 0 => Note that V= 0 is when the ball reached the maximum height.

V(7)= -96 ft/sec it means falling down

d)The ascending and descending times are equal, the ball is at maximum height when 4 seconds, which means it

will hit the ground after 8 seconds

Let, 128t - 16t^2 = 0

=> 16t(8t - t) = 0

t = 0 , 8 => (notice that at t = 0 the ball was thrown up ward from the ground)

t = 8=> the ball hits the ground after 8 second.

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