# Finding Position with a Quadratic Velocity

1. Sep 17, 2009

### glockstock

1. The problem statement, all variables and given/known data
In the time interval from 0.0 s to 10.8 s, the acceleration of a particle traveling in a straight line is given by ax = (0.1 m/s3)t. Let to the right be the +x direction. The particle initially has a velocity to the right of 10.0 m/s and is located 4.8 m to the left of the origin.

2. Relevant equations
(a) Determine the velocity as a function of time during the interval.

(b) Determine the position as a function of time during the interval.

(c) Determine the average velocity between t = 0.0 s and 10.8 s.
3 m/s

Compare it to the average of the instantaneous velocities at the start and ending times. Are these two averages equal?

3. The attempt at a solution

(a) v(t) = (0.05t^2)+10 this was determined to be right...

(b) x(t) = (0.017t^3)+10t+4.8 this is wrong...why? Significant Figures?

2. Sep 17, 2009

### rock.freak667

that means for t=0, x= -4.8m (+ve is to the right means -ve is to the left)