Ke Min is driving along on a rainy night at 23 m/s when he sees a tree branch lying across the road and slams on his breaks when the branch is 60.0 m in front of him. If the coefficiant of kinetic friction between the cars locked tires and the road is 0.41, will the car stop before hitting the branch? The car has a mass of 2400 kg.

### 2 Answers

From Newton second Law :

-0.4*m*g=ma

-0.4*g=a if g = 10m/s^2 the a=-4 m/s^2

using :

v=vo+at assume v=0 if the car stop

0 = 23-4*t then t= 5.75 s

using :

s=vo*t+0.5*a*t^2 = 132.25-0.5*4*5.75*5.75=132.25-66.125 = 66.125 m..

So, the car does not stop…the car hit the branch..

The question is asking whether he will stop within the 60 metres before the tree branch. This is a kinematic equation.

First, we need to figure out the force due braking, and hence the decelleration.

We know that for friction, F = mu * mg. In this case, mu is 0.41, so F = 0.41 * 2400 * 9.81 = 9653.04

F = m a, 9653.04 = 2400 * a, a = 9563.04 / 2400 = 4.0221 m/s ( or -4.0221 m/s technically, as he is decellerating).

You should be able to figure the rest out with kinematics from here on.