Derivatives : Displacement, Velocity and Acceleration

A car brakes in an emergency, the moment the brakes are first applied is t=0. the car's distance, s, from the point at which the brakes were first applied is given by:

S = -0.1t^3 + 20t + 0.1

(i) Differentiate s to find formulae for the velocity and acceleration.
(ii) Find the time at which the car comes to a stop.
(iii) How far does the car travels before it comes to a stop.
(iv) What does the mathematical model say will happen to the car after it comes to a stop? What will actually happen? Why are the two different?

Please see the attached file for the fully formatted problems.

Here is your solution (i've assumed the units are seconds and metres as they're not in the question, so if you know different please alter them) ...

Solution Summary

A car brakes in an emergency, the moment the brakes are first applied is t=0. the car's distance, s, from the point at which the brakes were first applied is given by:

S = -0.1t^3 + 20t + 0.1

A range of questions are solved related to this function (see below)
All solutions are provided with workings.

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