Acceleration, velcocity, and position

[talaroue]

Homework Statement

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.

Homework Equations

a=t²

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 equation 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?

 

[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.

 

[talaroue]

oh, that makes more sense! Thank you!