Question

Find the standard equation of a body moving with a constant acceleration a along a coordinate line. The following properties are known. See attachment.

Answer 1

s = v0t + (1/2)at^2

Answer 2

Was that it? o-o lol

Answer 3

ktklown, want explain how you got that? :|

Answer 4

Actually just because that's a standard equation from physics class. what you'd actually do is integrate v = v0 + at to get position

Answer 5

to be more specific, integrate v(t) = v0 + at with respect to t

Answer 6

integradt 1 to get v = v0 + at the expression for the velocity at time t
integrate again to get expression for s the distance

Answer 7

then use the constant s0 to get the "+C" constant of integration.

Answer 8

I'll try. lol

Answer 9

need more help or r u ok?

Answer 10

Umm, I'm trying to solve it. I integrated v(t) = v0 + at and got (at^2)/(2)+ tv0 +C

Answer 11

Am I on the right track? Or did I mess up? :|

Answer 12

no - thats fine so far

Answer 13

now u need to find value of C

Answer 14

Cool, integrate again?

Answer 15

Wait, no she doesn't have to integrate again, that's already the position equation

Answer 16

no - use the information you have in the question

Answer 17

you just have to solve for C now using the initial conditions given

Answer 18

yup

Answer 19

Okay, I make (at^2)/(2)+ tv0 +C equal tooo...s0?

Answer 20

Yes, at t = t0

Answer 21

Err, I mean, when t = 0. that's what your problem says.

Answer 22

s = 0 when t = 0

Answer 23

So, (at^2)/(2)+ tv0 +C= 0?

Answer 24

yes and since t = 0 then C = 0

Answer 25

You were closer earlier :-). you said: "Okay, I make (at^2)/(2)+ tv0 +C equal tooo...s0?"

Yes, you do that at t=0. in otherwords, substitute t=0 and solve for C.

Answer 26

So the constant is 0? Geezz... -____-;

Answer 27

right!!!!

Answer 28

so close! Your problem says that (at^2)/(2)+ tv0 +C = s0 when T=0. That reduces to C = s0.

Answer 29

u got there

Answer 30

jimmy it's not C=0, since there's an initial position s0

Answer 31

Wow, I hate these condition problems with a passion.

Answer 32

oh yes - sorry - i misread the question at t=0 s = s0 right

Answer 33

Aaaah, okaay. Hmm...I seee. Good job guys. :) Thanks.