kingjewel1 Posted January 29, 2006 Posted January 29, 2006 could you please help me with this question I would be very much obliged. I cant hack it for my exam A racing car of mass m travels along a straight road. While travelling the racing car is subject to a constant frictional resistance equal to ma (a is a positive constant) and aire resistance equal to k times thee square of its velocity v. The engine of the care can rpovide a constant propelling froce equal to mb (b is a positive constant) and can bring the car to a terminal velocity Vinfinity. a) given that the racing car starts from rest, show that it reaches a speed of Vinfinity/2 in time [math]\frac{v_{\infty}log3}{2(b-a)}[/math] b) at this point the engine is switched off. Show that the racing car comes to rest in a further time [math]\frac{v_{\infty}}{\sqrt(a(b-a))}[/math][math]arctan\sqrt\frac{b-a}{4a}[/math] thank you very much in advance
swansont Posted January 30, 2006 Posted January 30, 2006 F = m dv/dt = mb-kv2-ma You want to solve for vinfinity, the case where F = 0, and put that back into the equation. Then, for the second part, solve again setting mb to zero.
kingjewel1 Posted January 30, 2006 Author Posted January 30, 2006 Thank you swansont. I dont really understand the first part, could you explain it a bit more please.
swansont Posted January 30, 2006 Posted January 30, 2006 Force is mass x acceleration (F=ma), and a = dv/dt, i.e. the time-derivative of velocity. (using "a" as a constant confuses things, so I refrained from using F=ma in the same equation as where the constant appears) So you have a differential equation that needs to be solved, and you set dv/dt equal to the sum of all the forces; since it's a vector, you have to use the proper sign for the terms: + if they cause an acceleration in one direction, - if they cause an acceleration in the other direction. Then math happens. You know that at the terminal velocity the force is zero, which allows you to put the term with k in it in terms of vinfinity, which is what was done in the answer you gave.
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