# Math Help - Calculus 12 help! position functions

1. ## Calculus 12 help! position functions

Okay so Im in Calculus 12, and need to take a test tomorrow and am stuck on a question.
So.. the first part of the question is: A particle moves in a straight line with position function s(t), t>0. In each case find the velocity and acceleration functions, so i know velocity is first derivate and acceleration is second derivative
so the two equations are
2t^3-3t^2+6t and t^2- 4t+4
so velocities would be: 6t^2-6t+6 and 2t-4
acceleration would be : 12t - 6 and 2
Is that right???
Then heres the part I got stuck it says:
For both position functions answer the following
A) When is the particle at rest? (is this x intercepts?)
B)When is the particle moving in the positive and in the negative direction?
C) When is the speed of the particle increasing and when is it decreasing?
D) What is the total distance traveled after 6 sec of motion, starting at t=0

if someone could explain how im supposed to do those that would be so greatly appreciated
thanks!

2. Originally Posted by bek103
Okay so Im in Calculus 12, and need to take a test tomorrow and am stuck on a question.
So.. the first part of the question is: A particle moves in a straight line with position function s(t), t>0. In each case find the velocity and acceleration functions, so i know velocity is first derivate and acceleration is second derivative
so the two equations are
2t^3-3t^2+6t and t^2- 4t+4
so velocities would be: 6t^2-6t+6 and 2t-4
acceleration would be : 12t - 6 and 2
Is that right???
Then heres the part I got stuck it says:
For both position functions answer the following
A) When is the particle at rest? (is this x intercepts?) when v(t) = 0

B)When is the particle moving in the positive and in the negative direction?

the sign of v(t) indicates direction

C) When is the speed of the particle increasing and when is it decreasing?

speed increasing when v(t) and a(t) have the same sign

speed decreasing when v(t) and a(t) have opposite signs

D) What is the total distance traveled after 6 sec of motion, starting at t=0

total distance = $\textcolor{red}{\int_0^6 |v(t)| \, dt}$
...