Velocity question

[elle.123]

How do you tell whether v is meant to be positive or negative with questions like these?

Question: A particle initially at rest at an origin, moves along a straight line with a velocity v. The particle has an acceleration of (3+2x)^-2, where x is its position from the origin.

I have got to v= ±√((-1)/(3+2x)), and i don't know how to figure out if v is meant to be positive or negative.

TIA.

[Mao]

Initial acceleration is , this is in the positive x direction. Hence the object moves from origin in the positive x direction, :. v is positive.

However, I do believe your expression for v is incorrect:



, given

taking the positive branch.

[plato]

How do you tell whether v is meant to be positive or negative with questions like these?

Question: A particle initially at rest at an origin, moves along a straight line with a velocity v. The particle has an acceleration of (3+2x)^-2, where x is its position from the origin.

I have got to v= ±√((-1)/(3+2x)), and i don't know how to figure out if v is meant to be positive or negative.

TIA.

Your integral is not quite complete as you have forgotten the constant of integration. Without it, you would not get v=0 at x=o as stated in your question.

You should get
Then, substituting v=0 at x=0, you get c =
and so   

Then

From here, consider that the acceleration function is always positive (due to the squaring effect of the power) and tending toward a value of zero.
Therfore the velocity must be in the direction of the acceleration and tending toward a limiting value of as x gets larger.