1. ## integration problem

The position function x(t) of a particle moving along an x axis is x = 4.00 - 7.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? Graph x versus t for the range -5 s to +5 s. (e) To shift the curve rightward on the graph, should we include the term +20t (denote 1) or the term -20t (denote 0) in x(t)? (f) Does that inclusion increase (denote 1) or decrease (denote 0) the value of x at which the particle momentarily stops?

My teacher never got into this subject and i really need help

2. For part A) Take the derivative and set it equal to 0. $\displaystyle f'(x) = v(x)$ also know as the velocity funciton.
$\displaystyle x = 4.00 - 7.00t^2$

So you get:
$\displaystyle v(x) = -14t$
$\displaystyle 0 = -14t$
$\displaystyle t = 0$

So at time 0, the velocity is 0 (not moving).

For part B) To find out where, substitute 0 back in as t.
$\displaystyle x = 4 - 7(0)^2$
$\displaystyle x = 4$

So at a position of 4, the particle is not moving.