Posted by **Yoona** on Monday, November 7, 2011 at 9:07am.

A particle moves on the x –axis so that its position at any time is given by x(t) = 2t3 + 1.

a. Find the acceleration of the particle at t = 0.

b. Find the velocity of the particle when its acceleration is 0.

c. Find the total distance traveled by the particle from t = 0 to t = 5.

- calculous -
**Damon**, Monday, November 7, 2011 at 10:39am
x = 3 t^3 + 1

dx/dt = 9 t^2

d^2x/dt^2 = 18 t

so at t = 0 both acceleration and velocity are 0

velocity is always greater than 0 so the particle never backs up. Therefore all we need is the position at t = 5 minus the position at t = 1

x(5) = 251

x(1) = 7

------------

difference = 244

- calculous -
**Yoona**, Monday, November 7, 2011 at 12:18pm
thank you!!!!!!!!!!! I don't think any of my teachers would deticate some of their time for us:D

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