Question

If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2. Find the average velocity for the time period beginning when t = 2 and lasting for each of the following: 2.5 s, 2.1 s, 2.05 s, and 2.01 s.

Answer 1

The formula for average velocity between two times t1 and t2 of the position function f(x) is (f(t2)-f(t1)) / (t2-t1)
Plugging the values in for the first time period we get (f(2.5)-f(2)) / (2.5-2)
=> (f(2.5)-f(2)) / 0.5
f(2) will be the same for all 4 time periods and is
48(2)-16(2)^2 = 32
Now we plugin the other values
f(2.5) = 48(2.5)-16(2.5)^2 = 20
f(2.1) = 48(2.1)-16(2.1)^2 = 30.25
etc.
f(2.05) = 31.16
f(2.01) = 31.8384
Now plug these values into the formula
(20-32)/0.5 = -24
(30.25-32)/0.1 = -17.5
etc.
= -16.8
= -16.16
Final answer:
2.5s => -24 ft/s
2.1s => -17.5 ft/s
2.05 => -16.8 ft/s
2.01 => -16.16 ft/s
Hope I helped :)