help with this problem s'(t)

[motoxxrider33]

basically the problem is asking me to prove that on the graph s(t) = - 1/2at^2 +b with a and b being constants, that the average velocity over the time interval
[c - delta t, c + delta t] is equal to the instant velocity at t = c which is -ac - 1/2ah. but i dont understand the delta t. thats the little triangle right next to the t. haha sorry.
thanks for any time you devote to helping me

 

[daon]

Don't worry about the symbol "delta t." It merely means "a change in time." Its just a positive number, and you may use any symbol you'd like in your scratch work (say "p") until you feel more comfortable (I'd be sure to put the symbol back when you're handing in your work though ).

You're trying to show that \(\displaystyle \bar{V}[c-\Delta t, c+\Delta t] = -ac\)

The average velocity over the time interval [a,b] is

\(\displaystyle \bar V = \frac{S(b)-S(a)}{b-a}\)

In our case it means:

\(\displaystyle \bar V = \frac{S(c+\Delta t)-S(c-\Delta t)}{[c+\Delta t]-[c-\Delta t]} = \frac{[-\frac{1}{2}a{(c+\Delta t)}^2+b]-[-\frac{1}{2}a{(c-\Delta t)}^2+b]}{2\Delta t}\)

Simplifying, we get:

\(\displaystyle \frac{-2ac\Delta t}{2\Delta t} = -ac\)