- #1
- 3
- 0
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 dont need it worked out as much as I just need a push in the right direction. Thanks.
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 dont need it worked out as much as I just need a push in the right direction. Thanks.