Question

If a ball is thrown in the air with a velocity 52 ft/s, its height in feet t seconds later is given by y = 52t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting (i) 0.5 second. (ii) 0.1 second. (iii) 0.05 second. (iv) 0.01 second.

Answer 1

More generally, you have after
`h` seconds


`(y(2+h)-y(2))/((2+h)-2)=\frac{52(2+h)-16(2+h)^2-104-64}h=\frac{-12h-16h^2}h=-12-16h`

After
`h=0.5` second, the average velocity is
`-12-16*0.5=-20`.

I'm sure you can do the rest on your own.

Answer 2

(i)
`m=-20`

(ii)
`m=-13.6`

(iii)
`m=-12.8`

(iv)
`m=-12.16`

Explanation:

The given function is

`y=52t-16t^2`

where, h is the height of ball after t seconds.

It can be written as

`f(t)=52t-16t^2`

At x=2,

`f(2)=52(2)-16(2)^2=40`

The average of a function f(x) on [a,b] is

`m=(f(b)-f(a))/(b-a)`

The time period beginning when t = 2 and lasting with h.

`m=(f(2+h)-f(2))/(2+h-2)`

`m=(f(2+h)-40)/(h)`

(i)

Here, h = 0.5 second

`m=(f(2+0.5)-40)/(0.5)`

`m=(f(2.5)-40)/(0.5)`

`m=(30-40)/(0.5)`

`m=-20`

(ii)

Here, h = 0.1 second

`m=(f(2+0.1)-40)/(0.1)`

`m=(f(2.1)-40)/(0.1)`

`m=(38.64-40)/(0.1)`

`m=-13.6`

(iii)

Here, h = 0.05 second

`m=(f(2+0.05)-40)/(0.05)`

`m=(f(2.05)-40)/(0.05)`

`m=(39.36-40)/(0.05)`

`m=-12.8`

(iv)

Here, h = 0.01 second

`m=(f(2+0.1)-40)/(0.01)`

`m=(f(2.01)-40)/(0.01)`

`m=(39.8784-40)/(0.01)`

`m=-12.16`