When Does the Car Come to Rest Again?

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SUMMARY

The discussion centers on determining the time it takes for a car, described by the distance function x(t) = 2.30t² - 0.120t³, to come to rest after starting from rest. The user identifies key points in the car's motion, noting that the car reaches a maximum velocity of 14 m/s at t = 5 seconds and begins to decelerate thereafter. By taking the derivative of the distance function to obtain the velocity function and setting it to zero, the user successfully applies the quadratic formula to find the time at which the car returns to rest.

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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 = 2.30 m/s^2 and c = 0.120 m/s^3.

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


Homework Equations


dx/dt


The Attempt at a Solution


The problem I'm having with this question is I can't figure out how you can find the deceleration after the car stops accelerating. I know at t = 0 the velocity is 0 m/s, at t = 5s the velocity is 14m/s, and at t = 10s, the velocity is 10m/s this means the car decelerated from t = 5s to t = 10s. How do I find the deceleration to find how long it took the car to reach maximum velocity assuming it's 14.0m/s to come back to rest?
 
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Try using the fact that a=(delta v)/(delta t)
 
Ok, I got it...I just took the derivative of the distance formula which would make it into the velocity formula. Then I set the velocity equal to 0 and used the quadratic equation to find t.
 
Sounds right
 

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