Question

If a ball is thrown In the air with a velocity 48 ft/s. Its height in feet t seconds later is given by y =48t-16t2. (Round your answers to one decimal place.)
a. Find the average velocity for the time period beginning when t = 2 and lasting 0.5 second.
b. Find the average velocity for the time period beginning when t = 2 and lasting 0.05 second.
c. Find the average velocity for the time period beginning when t = 2 and lasting 0.01 second.
d. Estimate the instantaneous veiocity when t=2.

Answer 1

Answer:

(a)-24 ft/s

(b)-16.8 ft/s

(c)-16.16 ft/s

(d) -16 ft/s

Explanation:

The initial velocity, v=48 ft/s

(a)

The average velocity when t=2 and 0.5 seconds later hence t=2.5 is given by

$\frac {f(2)-f(2.5)}{2-2.5}=\frac {[48(2)-16(2)^{2}]-[48(2.5)-16(2.5)^{2}]}{2-2.5}=\frac {32-20}{-0.5}=-24 ft/s$

(b)

The average velocity when t lasts 0.05 second then

t=2 and t=2.05 is given by

$\frac {f(2)-f(2.05)}{2-2.05}=\frac {[48(2)-16(2)^{2}]-[48(2.05)-16(2.05)^{2}]}{2-2.5}=\frac {32-31.16}{-0.05}=-16.8 ft/s$

(c)

The average velocity when t=2 and lasts 0.01 s then t=2.01 then is determined by

$\frac {f(2)-f(2.01)}{2-2.01}=\frac {[48(2)-16(2)^{2}]-[48(2.01)-16(2.01)^{2}]}{2-2.01}=\frac {32-31.8384}{-0.01}=-16.16 ft/s$

(d)

As the intervals get small, that is to imply from 0.05 s later to 0.01 s later as seen in parts a to c, the values are getting small and closer to 16 hence the instantaneous velocity at t=2 is -16 ft/s