Question

A girl 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.

Answer 1

To find when the velocity is 0 m/s, differentiate the position function to obtain velocity, and solve for when it is zero. For the position and total distance when acceleration is zero, differentiate velocity to obtain acceleration, find when it is zero, then calculate position and integrate velocity over time for distance.

Step-by-step explanation:

To determine when the velocity is 0 m/s for the radio-controlled model car, we need to find the first derivative of the position function x(t), which represents velocity v(t). We differentiate x(t) = 0.5t3 - 3t2 + 3t + 2 with respect to time t, yielding v(t) = 1.5t2 - 6t + 3. Setting v(t) = 0 and solving for t will give us the times at which the velocity is zero.

To find the position and the total distance travelled when the acceleration is zero, we need to find the second derivative of the position function x(t), which represents acceleration a(t). We differentiate v(t) = 1.5t2 - 6t + 3 with respect to time t, yielding a(t) = 3t - 6. Setting a(t) = 0 and solving for t will give us the time at which the acceleration is zero. Then we substitute that time back into x(t) to get the position, and into v(t) to integrate for total distance travelled up to that time, considering any changes in direction that might require us to take the absolute value of distances travelled during intervals.

Answer 2

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.

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

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