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.

More generally, you have after [tex]h[/tex] seconds

[tex]\dfrac{y(2+h)-y(2)}{(2+h)-2}=\dfrac{52(2+h)-16(2+h)^2-104-64}h=\dfrac{-12h-16h^2}h=-12-16h[/tex]

After [tex]h=0.5[/tex] second, the average velocity is [tex]-12-16\times0.5=-20[/tex].

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

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erinna erinna · Answer 2 · 2019-10-03T10:48:09+02:00

Answer:

(i) [tex]m=-20[/tex]

(ii) [tex]m=-13.6[/tex]

(iii) [tex]m=-12.8[/tex]

(iv) [tex]m=-12.16[/tex]

Step-by-step explanation:

The given function is

[tex]y=52t-16t^2[/tex]

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

It can be written as

[tex]f(t)=52t-16t^2[/tex]

At x=2,

[tex]f(2)=52(2)-16(2)^2=40[/tex]

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

[tex]m=\frac{f(b)-f(a)}{b-a}[/tex]

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

[tex]m=\frac{f(2+h)-f(2)}{2+h-2}[/tex]

[tex]m=\frac{f(2+h)-40}{h}[/tex]

(i)

Here, h = 0.5 second

[tex]m=\frac{f(2+0.5)-40}{0.5}[/tex]

[tex]m=\frac{f(2.5)-40}{0.5}[/tex]

[tex]m=\frac{30-40}{0.5}[/tex]

[tex]m=-20[/tex]

(ii)

Here, h = 0.1 second

[tex]m=\frac{f(2+0.1)-40}{0.1}[/tex]

[tex]m=\frac{f(2.1)-40}{0.1}[/tex]

[tex]m=\frac{38.64-40}{0.1}[/tex]

[tex]m=-13.6[/tex]

(iii)

Here, h = 0.05 second

[tex]m=\frac{f(2+0.05)-40}{0.05}[/tex]

[tex]m=\frac{f(2.05)-40}{0.05}[/tex]

[tex]m=\frac{39.36-40}{0.05}[/tex]

[tex]m=-12.8[/tex]

(iv)

Here, h = 0.01 second

[tex]m=\frac{f(2+0.1)-40}{0.01}[/tex]

[tex]m=\frac{f(2.01)-40}{0.01}[/tex]

[tex]m=\frac{39.8784-40}{0.01}[/tex]

[tex]m=-12.16[/tex]