Question

Can you just pls tell me the answer first then explain it ?

Answer 1

Part (a)

The frictional force acting on the car can be given as,

$f=-\mu mg$

According to Newton's law, the net force acting on the car is,

$F=ma$

At the equilibrium,

$F=f$

Plug in the known expressions,

$\begin{gathered} ma=-\mu mg \\ a=-\mu g \end{gathered}$

The final speed of the car can be given as,

$v^2=u^2+2ad$

Substitute the known expression,

$v^2=u^2+2(-\mu g)d$

Substitute the known values,

$\begin{gathered} (0m/s)^2=(52.4km/h)^2(\frac{0.278\text{ m/s}}{1\text{ km/h}})^2-2(0.123)(9.8m/s^2)d \\ d=\frac{212.2m^2s^{-2}}{2.41m/s^2} \\ \approx88.0\text{ m} \end{gathered}$

Thus, the minimum distance travelled by car to stop is 88.0 m.

Part (b)

Now, the coefficient of friction is 0.594.

Substitute the known values,

$\begin{gathered} (0m/s)^2=(52.4km/h)^2(\frac{0.278\text{ m/s}}{1\text{ km/h}})^2-2(0.594)(9.8m/s^2)d \\ d=\frac{212.2m^2s^{-2}}{11.64m/s^2} \\ \approx18.2\text{ m} \end{gathered}$

Thus, the stopping distance travelled of the car is 18.2 m.