Question

As a certain object falls, its position s (in meters) above ground after t seconds is given by s(t)=20-5t^2. Whats the average velocity of the object on the interval from t=1 to the time .5 seconds later?? from t=1 to the time .1 seconds later?

Answer 1

Why did you delete your previous post? I almost finished there -.- !!
Anyway, the average velocity is equal to the displacement divided by the total time. In this case, the average velocity \(=\frac{s(1.5)-s(1)}{0.5}=\frac{8.75-15}{0.5}=-12.5 m/s\).

Answer 2

You can similarly do the second part of the problem.

Answer 3

v(t) = -10t ===> v(1) = -10 , v(1.5) -=10(1.5) = -15 ==> average velocity = [-15 -10]/0.5 = -12.5

Answer 4

whoops sorry!! :/ thank you though!

Answer 5

No problem :)