Question

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: $\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v\frac{dv}{dx}$.)

Answer 1

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, $\mu_k=0.5$

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

$-f=ma$

$a=\dfrac{-f}{m}$

$a=\dfrac{-\mu _k mg}{m}$

$a=\mu_k g$

$a=-0.5\times 9.8$

$a=-4.9\ m/s^2$

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

$v.dv=a.dx$, x is the stopping distance

$\int\limits^0_v {v.dv}=\int\limits^x_0 {a.dx}$

$\dfrac{v^2}{2}|_v^0=ax$

$0-(27.77)^2=-2\times 4.9x$

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.