# Kinematics Problem and final velocity

Brett
Problem: The acceleration of a particle is given by Ax(t) = -2.00m/s^2 + (3.00m/s^2)t. A) Find the initial velocity such that the particle will have the same x-coordinate at t= 4.00s as it had at t= 0. B) What will the velocity be at t= 4.00s?

Work so far:

Integrated to get these:

Vx(t) = -2t + (3/2)t^2
x(t) = -t^2 + (1/2)t^3

I am just stuck on what to do. I don't need it worked out as much as I just need a push in the right direction. Thanks.

PureEnergy
Both your equations are wrong. When you integrated the acceleration, you forgot about the constant of integration or the initial velocity.

Brett
PureEnergy said:
Both your equations are wrong. When you integrated the acceleration, you forgot about the constant of integration or the initial velocity.

So then they are:

Vx(t) = Vi + [-2t + (3/2)t^2]
x(t) = Xi + Vi(t) + [-t^2 + (1/2)t^3]

With Vi meaning initial velocity and Xi meaning initial position.

I just solved it, thanks for the correction.

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