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- September 19th 2009, 07:22 PM #1

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## Dv/dt=-2e^-x v=2e^-(x/2) anyone willing to try this one?

Hi,

I am trying to figure out this question but I really can't understand it. Can someone please work through it and explain in a simple way what the steps are?

A particle is moving on a straight line path with acceleration given by dv/dt=-2e^-x where x is the distance of the particle from the origin and v is the velocity of the particle.

Knowing that d/dx(v^2/2)=dv/dt,hence determine that if v=2 when x=0 then v=2e^-(x/2)

Thanks

David

- September 19th 2009, 07:31 PM #2

- September 19th 2009, 10:10 PM #3

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OK, correct me if I'm wrong which I probably am,

v(x) is the indefinate integral of dv/dt=-2e^-x which would be v=2e^-x+c if that is the case then I'm not exactly sure what the method would be, but I'll try anyway. If x=0 the e^-x=1 so for 2*1+c to equal 2, c must be 0 so v=2e^-x? Was I close?

p.s. sorry about the reply being such a mess, I'm totally new to LaTex (like yesterday) and I'm not on top of it.

David

- September 19th 2009, 10:23 PM #4