Question

If the position of a particle on the x-axis at time t is s(t)=-5^2, then the average velocity of the particle for 0<=t<=3 is?

Answer 1

\[\frac{f(3)-f(0)}{3-0}\] i think

Answer 2

i bet it is not really \(s(t)=5^2\) though...

Answer 3

velocity is \(\large\color{slate}{ s'(t) }\)

Answer 4

wait, your s(t) is likely with a typo

Answer 5

oops s(t)=-5t^2

Answer 6

yes, so s'(t)=?

Answer 7

i am pretty sure it is asking for average velocity on the interval not instantaneous velocity
i could be wrong i suppose

Answer 8

\(\large\color{slate}{ s (t) =-5t^2 }\)
\(\large\color{slate}{ s '(t) =? }\)

Answer 9

-10t

Answer 10

yes, \(\large\color{slate}{ s'(t) =-10t }\)

Answer 11

try
\[\frac{-5\times 3^2-(-5\times 0^2)}{3-0}\]

Answer 12

the derivative s'(t) is velocity, not s(t). s9t) is the position

Answer 13

then the AVERAGE velocity of the particle

Answer 14

it would be the average derivative from 0 to 3

Answer 15

the average derivative?

Answer 16

derivative is the velocity, and you are finding the average velocity.... don't you think so?

Answer 17

I have to leave now... apologize guys

Answer 18

if you find the average of the derivative over that interval, it is an integral
the integral will be the original function
evaluate at the endpoints, divide by the length of the interval, it is identical to
\[\frac{s(3)-s(0)}{3-0}\]

Answer 19

-15

Answer 20

\[\frac{-5\times 3^2}{3}=-15\] looks good to me
why not?

Answer 21

:)

Answer 22

it was \(s(t)=-5t^2\) right?

Answer 23

well in fact solomon was right
just unnecessary
if you take the average value of the derivative, you get the same thing

Answer 24

yes :)

Answer 25

Yup. I was about to argue that there's no need to involve calculus in this because the definition of average velocity is \(\Delta s/\Delta t\) -- nothing instantaneous in that... which is also what you said, so...