# Determining initial velocity and instant velocity

1. Jan 22, 2010

### Norfonz

1. The problem statement, all variables and given/known data

The acceleration of a particle is given by ax(t) = 2.90t - 2.06

Q. Find the initial velocity v0x such that the particle will have the same x-coordinate at time t = 4.06 as it had at t = 0.

2. Relevant equations

vx = v0x + $$\int$$axdt evaluated from 0 to t.
x = x0 + $$\int$$vxdt evaluated from 0 to t.

This is defined for straight-line motion with varying acceleration, which appears to fit my scenario.

3. The attempt at a solution

vx = v0x + $$\int$$(2.90t - 2.06)dt evaluated from 0 to t.
vx = v0x + 1.45t2 - 1.03t

x = x0 + $$\int$$(v0x + 1.45t2 - 1.03t)dt evaluated from 0 to t.
x = x0 + $$\int$$(1.45t2 - 1.03t)dt + v0x$$\int$$dt

x = x0 + v0xt + (1.45t3/3) - (1.03t2/2)

Unfortunately this is as far as I got. I don't understand how to incorporate the values I was given for t. Could someone point me in the right direction please?

2. Jan 22, 2010

### rock.freak667

Well from where, when t=4.06, what is x equal to ? (in terms of v0x and x0)

And when t=0, what is x equal to ? (in terms of v0x and x0)

Then just put those equal to each other and solve for v0x