Motion in a straight line (1 Viewer)

[swagswagyoloswag]

The acceleration of a particle is 2x-5 m/s^(2) where x is the distance in metres from the origin.
a) Find expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially.
I got v^2 = 2x^2 - 10x -12 but can't find a condition to just find v. Also, when I let v = 0, it shows that it stops at two positions, x = -1 and x = 6.
But since it starts at rest at x = -1, a = 2(-1)-5 = -7 initially, wouldn't it mean it only travels left because v = 0, a<0 ?

 

[InteGrand]

The acceleration of a particle is 2x-5 m/s^(2) where x is the distance in metres from the origin.
a) Find expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially.
I got v^2 = 2x^2 - 10x -12 but can't find a condition to just find v. Also, when I let v = 0, it shows that it stops at two positions, x = -1 and x = 6.
But since it starts at rest at x = -1, a = 2(-1)-5 = -7 initially, wouldn't it mean it only travels left because v = 0, a<0 ?

The question's saying that x is the distance from the origin. They mean displacement, right?

 

[swagswagyoloswag]

The question's saying that x is the distance from the origin. They mean displacement, right?

yep itis

 

[InteGrand]

yep itis

OK

 

[InteGrand]

Then yes, x will have to stay negative, since initially x < 0, v = 0 and and a < 0, so the particle will start moving to the left, and the more left it moves, the more negative the acceleration becomes, meaning it keeps going left (and gets faster). So you can discard any solutions with positive x.

 

[swagswagyoloswag]

ok thanks