# Homework Help: Acceleration, velcocity, and position

1. Jan 7, 2010

### talaroue

1. The problem statement, all variables and given/known data
The acceleration of a particle is directl porportional to the square of the time t. When t=0s, and the particle is at x=24m. Knowing that at t=6s, x=96m, and V=18m/s, express x and v in terms of t.

2. Relevant equations

a=t2

3. The attempt at a solution

so i took the integral of a=t2 and got v=1/3t3+C, then took the integral again to obtain x=1/12t4+Ct+A.

So then I plugged in x=24 and t=0 to get A=24

so now my eqation is.....
x=1/12t4+Ct+24 solving for C i get C=-6

so then my equation is x=1/12t4-6t+24 in the book the answer is 1/108t4+10t+24

what am I doing wrong?

2. Jan 7, 2010

### ehild

Directly proportional does not mean equal. So a = kt2, where k is a unknown constant. You get it from the data given for t=0 and t=6 s.

3. Jan 7, 2010

### talaroue

oh, that makes more sense! Thank you!