For a 1,582-kg car with a drag constant D=0.49 kg/m, which is moving at a speed

Question

For a 1,582-kg car with a drag constant D=0.49 kg/m, which is moving at a speed of 39 mph, what is the magnitude of the maximum acceleration (right at this speed) at which the car can further speed up. The coeffcicient of kinetic friction is 0.55 and the coefficient of static friction is 0.79.

Calculate the answer to at least 3 significant figures (this is too accurate for practical purposes, but will test whether you really understand this problem.)

(Note, this will be the limit due to friction; the real maximum might be further limited by engine power or torque.)

Explanation / Answer

Here , v = 39 mph = 17.43 m/s

as D = 0.49 Kg/m

Drag force = 0.5*rho*v^2*D

Fd = 0.5*1.214 * 0.49 * 17.34^2

Fd = 90.36

Now, acceleration due to drag

ad = - Fd/m

ad = -90.36/1582

ad = -0.057 m/s^2

Now,

maximum acceleration ,

Amax = us*g - ad

Amax = 0.79*9.8 - .057

Amax = 7.68 m/s^2

magnitude of the maximum acceleration is 7.68 m/s^2

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