Question

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Answer 1

That integration looks correct since it's being integrated with respect to t. Don't forget the +C

Answer 2

so x is the distance m1 is from m2?

Answer 3

if so, then how is m2 equal to x?

Answer 4

What's the full initial problem?

Answer 5

How did you define x?

Answer 6

so it's dv/dx instead of dv/dt?

I think it should be dv/dt because acceleration is change in velocity over change in time

Answer 7

so your integration will be with respect to t, not x

Answer 8

well dx/dt represents the instantaneous rate of change of the particles x position with respect to time

Answer 9

ie

Vx = dx/dt

Answer 10

that's the thing, I'm reading about the Runge-Kutta method and it's still very strange to me

so I can't help you with that part sadly

Answer 11

you mean break it down into x and y components?

Answer 12

i mean i think i'm supposed to be able to just use the right hand side of dv/dt which is the -GM1 stuff and im willing to try it but since my equation is with respect to "t", do i just turn the x's into t's?

Answer 13

basically i think the horizontal axis is t and the vertical is x in this case

Answer 14

t is time

x is the distance the particle M1 is away from M2. So basically the positive distance on the x axis (from 0 to M2). So I think x is still horizontal

Answer 15

since the problem is one dimensional...i believe x should be on the vertical axis but im not sure anymore...ugh

Answer 16

yeah I'm not sure of much of this problem either

Answer 17

you might be right though

Answer 18

oh well thanks jim

Answer 19

sorry I wasn't of much help

Answer 20

other people might know though