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Derivatives : Displacement, Velocity and Acceleration

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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?

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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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Hi there :)

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) ...

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