Question

If v(t)=-t^2+4t+8 and s(2)=20 find the displacement s(t) and any point in time?

Answer 1

Do I find the antiderivative of v(t) and then plug in s(2) to find the constant? I'm not sure.

Answer 2

hint:
\[\Large s(t) = \int v(t)dt\]

Answer 3

what is the antiderivative of v(t)?

Answer 4

I got s(t)=-(x^3/3)+(2x^2)+(8x)+C .. Would I plug into s(2) into this equation?

Answer 5

`I got s(t)=-(x^3/3)+(2x^2)+(8x)+C `
replace every x with t

Answer 6

\[\Large s(t) = -\frac{1}{3}t^3+2t^2+8t+C\]

Answer 7

plug in t = 2 and that will have s(2) on the left side

use the fact that s(2) = 20 to replace the "s(2)" with "20"

then solve for C

Answer 8

Then s(t)=(-t^3/3)+2t^2+8t-1.33 would be the answer?

Answer 9

C = -1.3333333333333... = -4/3

Answer 10

So,
\[\Large s(t) = -\frac{1}{3}t^3+2t^2+8t-\frac{4}{3}\]

Answer 11

Ah yes I should have put it in fraction form - thank you very much I was a little confused.

Answer 12

I'm glad I could help out