Acceleration of a particle

[Tucta]

A particle is moving on a straight line path with acceleration given by:

. . . . .$\displaystyle \dfrac{dv}{dt}\, =\, -2\, e^{-x}$

where x is the distance of the particle from the origin and v is the velocity of the particle.

Knowing that:

. . . . .$\displaystyle \dfrac{d}{dx}\, \left(\dfrac{v^2}{2}\right)\, =\, \dfrac{dv}{dt}$

hence determine that if v = 2 when x = 0 then:

. . . . .$\displaystyle v\, =\, 2\, e^{-x/2}$

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I'm at a complete loss, don't understand what the question is asking me to do.

 

[stapel]

A particle is moving on a straight line path with acceleration given by:

. . . . .$\displaystyle \dfrac{dv}{dt}\, =\, -2\, e^{-x}$

where x is the distance of the particle from the origin and v is the velocity of the particle.

Knowing that:

. . . . .$\displaystyle \dfrac{d}{dx}\, \left(\dfrac{v^2}{2}\right)\, =\, \dfrac{dv}{dt}$

hence determine that if v = 2 when x = 0 then:

. . . . .$\displaystyle v\, =\, 2\, e^{-x/2}$

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I'm at a complete loss, don't understand what the question is asking me to do.

You are being asked to show that the velocity "v" can be expressed in a certain way.

You have been given two different expressions for dv/dt. What you learned back in algebra suggests an equation you can make with this. Where does this equation lead?

 

[Deleted member 4993]

A particle is moving on a straight line path with acceleration given by:

. . . . .$\displaystyle \dfrac{dv}{dt}\, =\, -2\, e^{-x}$

where x is the distance of the particle from the origin and v is the velocity of the particle.

Knowing that:

. . . . .$\displaystyle \dfrac{d}{dx}\, \left(\dfrac{v^2}{2}\right)\, =\, \dfrac{dv}{dt}$

hence determine that if v = 2 when x = 0 then:

. . . . .$\displaystyle v\, =\, 2\, e^{-x/2}$

---

I'm at a complete loss, don't understand what the question is asking me to do.

.

$\displaystyle \frac{d}{dx}\left[\frac{v^2}{2}\right] \ = -2*e^{-x}$

$\displaystyle d\left[\frac{v^2}{2}\right ] \ = -2*e^{-x} \ dx$

Now integrate both sides - and you are done.....

 

[Steven G]

A particle is moving on a straight line path with acceleration given by:

. . . . .$\displaystyle \dfrac{dv}{dt}\, =\, -2\, e^{-x}$

where x is the distance of the particle from the origin and v is the velocity of the particle.

Knowing that:

. . . . .$\displaystyle \dfrac{d}{dx}\, \left(\dfrac{v^2}{2}\right)\, =\, \dfrac{dv}{dt}$

hence determine that if v = 2 when x = 0 then:

. . . . .$\displaystyle v\, =\, 2\, e^{-x/2}$

---

I'm at a complete loss, don't understand what the question is asking me to do.

I realized that dv/dt = d(2e^-x +c)/dx