Agirl operates a radio-controlled model car in a vacant parking lot. the girl’s position is at the origin of the xy coordinate axes, and the surface of the parking lot lies in the xy plane. she drives the car in a straight line so that the x coordinate is defined by the relation x(t) = 0.5t3 - 3t2 + 3t + 2 where x and t are expressed in meters and seconds, respectively. determine when the velocity is 0 m/s, and the position and the total distance travelled when the acceleration is zero.

The position at time t is
x(t) = 0.5t³ - 3t² + 3t + 2 

When the velocity is zero, the derivative of x with respect to t is zero. That is,
x' = 1.5t² - 6t  + 3 = 0
or
t² - 4t + 2 = 0

Solve with the quadratic formula.
t = (1/2) [ 4 +/- √(16 - 8)] = 3.4142 or 0.5858 s

When t =0.5858 s, the position is
x = 0.5(0.5858³) - 3(0.5858²) + 3(0.5858) + 2 = 2.828 m
When t=3.4142 s, the position is
x = 0.5(3.4142³) - 3(3.4142²) + 3(3.4142) + 2 = -2.828 m
Reject the negative answer.

The velocity is zero when t = 0.586 s, and the distance is 2.83 m

When the acceleration is zero, the second derivative of x with respect to t is zero. That is,
3t - 6 = 0
t = 2
The distance traveled is
x = 0.5(2³) - 3(2²) + 3(2) + 2 = 0

When the acceleration is zero, t = 2 s, and the distance traveled is zero.

Arise 8 run 1 this means go 8 up and 1 across, which is steepest