A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. (Hint: since the distance traveled is of interest rather than the time, x is the desired independent variable and not t. Use the Chain Rule to change the variable: [tex]\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v\frac{dv}{dx}[/tex].)

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Muscardinus Muscardinus · Answer 1 · 2019-10-16T08:40:30+02:00

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Speed of the car, v = 100 km/h = 27.77 m/s

The coefficient of kinetic friction of the tires, [tex]\mu_k=0.5[/tex]

Let f is the net force acting on the body due to frictional force, such that,

[tex]-f=ma[/tex]

[tex]a=\dfrac{-f}{m}[/tex]

[tex]a=\dfrac{-\mu _k mg}{m}[/tex]

[tex]a=\mu_k g[/tex]

[tex]a=-0.5\times 9.8[/tex]

[tex]a=-4.9\ m/s^2[/tex]

We know that the acceleration of the car in calculus is given by :

[tex]v.dv=a.dx[/tex], x is the stopping distance

[tex]\int\limits^0_v {v.dv}=\int\limits^x_0 {a.dx}[/tex]

[tex]\dfrac{v^2}{2}|_v^0=ax[/tex]

[tex]0-(27.77)^2=-2\times 4.9x[/tex]

On solving the above equation, we get, x = 78.69 meters

So, the stopping distance for the car is 78.69 meters. Hence, this is the required solution.