Question

an object is thrown from the top of a building that is 525 feet tall. the distance is from the ground after t seconds is:

s(t)=561-(4t-6)^2

a)Find the average velocity in the interval [2,4].
I got -48. Is that right?
b)Find the instantaneous velocity at t=4 seconds.
I am stuck on this one. I have
[561-(4t+4h-6)^2]/h

Can I cancel the h's out? Or does the square keep me from doing that? Once i get rid of the h do I just plug 4 in and that is my answer?

Answer 1

that is not the right thing. you need
\[\frac{s(t+h)-s(t)}{h}\]

Answer 2

you wrote out
\[s(t+h)\] but now you have to subtract
\[s(t)\]

Answer 3

that is the equation i started out with for b

Answer 4

should be
\[\frac{561-(4(t+h)-6)^2-(561-(4t-6)^2)}{h}\]

Answer 5

when you simplify in the numerator everything without in h in it will add up to zero, then you can cancel the h

Answer 6

what was the original formula before you replaced t by 4?

Answer 7

i replaced t with t+h.. i forgot to subtract s(t).

Answer 8

and you should replace t by 4

Answer 9

this is confusing.. oh you are not writing those. ha. ok so after i cancel things out in the numerator i have 4h/h?

Answer 10

and the answer is 4?