# Particle motion problem

1. Apr 7, 2007

### spin360

1. The problem statement, all variables and given/known data
A particle moving along the x-axis has its velocity described by the function vx = 5t^2 m/s, where t is in s. Its initial position is x0 = 1 m at t0 = 0 s. At t = 1 s, find the position, velocity, and acceleration of the particle.

So from that equation... vx = 5t^2, x0 = 1m, t0 = 0s, t1 = 1s

2. Relevant equations
v1 = v0 + at
x1 = x0 + v0 + (1/2)at^2

3. The attempt at a solution
So for the velocity, I just plugged in 1s for t and got 5.0m/s, which is correct. I'm stuck on the acceleration and position though, and I'm thinking it's because I don't know what v0 is. I assumed v0 = 0 because don't you plug in 0 for t in the given function? Which then the answer is 0. Any thoughts?

2. Apr 7, 2007

### Hootenanny

Staff Emeritus
Your right, the intial velocity is zero. You can now find the position and acceleration by noting that;

$$a = \frac{dv}{dt}\hspace{1cm}\text{and}\hspace{1cm}x = \int v dt$$

3. Apr 7, 2007

### exec

When velocity does not vary linearly with time, acceleration is not constant, which means, those equations which hold only when acceleration is constant cannot be used. In this case, velocity varies with the square of time, plotting a graph of velocity versus time shows that the graph is a quadratic function, not linear.

For the position and velocity part, I think you can make it through. I don't know if you've learned basic calculus before but velocity v = dx/dt(change of position with respect to time) and a = dv/dt(change of velocity with respect to time). Differentiate the v(t) function and plug in t = 1s and you get the acceleration.

4. Apr 7, 2007

### spin360

Ahh okay that makes sense. So I got the acceleration right, just by the derivative. I took the integral of 5t^2... which came out to (5t^3)/3... which gives me 1.667m for position. Webassign says that's incorrect?

5. Apr 7, 2007

### spin360

Nevermind, I forgot to add 1m :tongue: