Calculating Velocity and Acceleration for a Moving Car

[CursedAntagonis]

Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)= bt^2 - ct^3, where b = 2.90 m/s^2 and c = 0.150 m/s^3.

How long after starting from rest is the car again at rest?

Homework Equations

Already found

Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s which is 14 m/s.

Calculate the instantaneous velocity of the car at t=0 which is 0.

Calculate the instantaneous velocity of the car at t=5.00 s which is v = 17.8 m/s.

Calculate the instantaneous velocity of the car at t=10.0 s which is v = 13.0 m/s.

The Attempt at a Solution

I tried to find the acceleration from v = 17.8 m/s to 13.0 m/s which I got -0.96 but I don't know what equation I need to use to find the time for the car to come to a stop. I don't even know if I am on the right track.

 

[Dick]

You are given that the car begins at rest and ends at rest. So v(t)=0 at both those times. v(t)=x(t)'. If you set the equation for v(t)=0 you will find exactly two times when the car is at rest. What are they? I think you are doing fine on the other part of the problem.

 

[CursedAntagonis]

Ok so I got the equation 2bt - 3ct^2

I plugged in the given values and worked it out.

I got two values for t which are : 0, 12.8888889

I am assuming the 12.9 is the answer.

Am I correct Sir?

Also, thank you very much for the help!

 

[Dick]

That's the same thing I got. Don't mention it.