Question

A car is traveling at 52.4 km/h on a flat
highway.
The acceleration of gravity is 9.81 m/s2 .
a) If the coefficient of kinetic friction be-
tween the road and the tires on a rainy day is
0.137, what is the minimum distance needed
for the car to stop?
Answer in units of m

Answer 1

\[W=\Delta KE\] Work is also equal to a force over a distance, so \[Fd=\Delta KE\]
\[-Fd=1/2mv _{2}^{2} - 1/2mv _{1}^{2}\]
So we manipulate to get distance alone:
\[d=(1/2mv _{2}^{2} - 1/2mv _{1}^{2})/-F\]
We know that Ffric=kFnatural and Fnatural=mg so doing a replacement we get:
\[d=(1/2mv _{2}^{2} - 1/2mv _{1}^{2})/-(k*mg)\]
The m cancels out, final velocity is 0, we are left with:
\[d=(- 1/2v _{1}^{2})/-(k*g)\]
and after converting km/hr to m/s we have:
\[d=(- 1/2\times(14.6m/s)^{2})/-(0.137\times9.8m/s)\]
Giving you a stopping distance of 78.9m