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.
(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

y=48t -16t^2

Answer 2

\[y=48t-16t^2\] at at \(t=2\) you have \(48\times 2-16\times 2^2=32\) feet high

Answer 3

\(.5\) seconds later, \(t=2.5\) so we have to replace \(t\) by \(2.5\) then subtract \(32\) and then divide by \(.5\)
best to use a calculator, i will post my result

Answer 4

oh okay got it

Answer 5

http://www.wolframalpha.com/input/?i=%28+48*2.5+%E2%88%92+16%282.5%29^2-32%29%2F.5
i get \(-24\) so it is going down

Answer 6

using the link i wrote as a template, we can do the rest quick
\[\frac{48\times 2.1-16\times 2.1^2-32}{.1}\] for the next one

Answer 7

http://www.wolframalpha.com/input/?i=%28+48*2.1+%E2%88%92+16%282.1%29^2-32%29%2F.1

Answer 8

hope it is clear, and clear how i made the changes in wolfram to give the answer quickly

Answer 9

at 0.5 its at 12ft/s

Answer 10

that is not what i got

Answer 11

hold on
make sure you understand what the question is asking
it is not asking for the height at \(t=.5\)
it is asking for the rate of changes starting at \(t=2\) and ending \(.5\) seconds later, meaning at \(t=2.5\)

Answer 12

or so at 2.5 its at
24ft/s

Answer 13

-24 okay i got it

Answer 14

at \(t=2.5\) it is at 20 right?
so the rate of change is
\[\frac{20-32}{2.5-2}=\frac{-12}{.5}=-24\]