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

**Splint** · September 19th 2009, 07:22 PM

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

**VonNemo19** · September 19th 2009, 07:31 PM

Originally Posted by Splint

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

Your are asked to find a particular solution to this differential equation.

All your problem is saying is that

Given that $\frac{dv}{dt}=-2e^{-x}$ and $v(0)=2$

Find $v(x)$ ...

Can you do this?

**Splint** · September 19th 2009, 10:10 PM

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

**VonNemo19** · September 19th 2009, 10:23 PM

Originally Posted by Splint

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

Great job

PS, the easiest way to learn latex is to just click on someones latex that has been written. Click on the latex below:

$v(x)=\int-2e^{-x}=2e^{-x}+C$

$v(0)=2\Rightarrow{C}=0$

Therefore $v(x)=2e^{-x}$

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