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, xx is the desired independent variable and not tt. Use the Chain Rule to change the variable: \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx} ​dt ​ ​dv ​​ = ​dx ​ ​dv ​​ ​dt ​ ​dx ​​ =v ​dx ​ ​dv ​​ .)

Answer 1

Answer:

$s=78.7352\ m$

Explanation:

Given:

• mass of car, $m=1000\ kg$

• initial velocity of car, $u=100\ km.hr^{-1}=27.78\ m.s^{-1}$

• coefficient of kinetic friction, $\mu=0.5$

Now the kinetic frictional force:

$f=\mu.N$

where:

N= normal force of reaction by the ground on the car, which is equal to the weight of the car

$f=0.5\times (1000\times 9.8)$

$f=4900\ N$

Therefore the acceleration of the car due to frictional force:

$a=\frac{f}{m}$

$a=\frac{4900}{1000}$

$a=-4.9\ m.s^{-1}$

Now the stopping distance:

$v^2=u^2+2a.s$

$0^2=27.78^2+2\times (-4.9)\times s$

$s=78.7352\ m$