Formulas:

v=d/t
vf* - vi*=2ad Ff=MFn

d=1/2at* + vt F=ma Fn=mg

x=-b+- (squareroot of) [b-4ac] / 2a

symbols:

*=squared M=coefficent of friction v=velocity(m/s) d=displacement(m)
t=time (s) vf=final velocity vi=initial velocity

a=acceleration(m/s*) F=force (Newtons) m=mass (kg)

g=force of gravity (9.81 m/s*) Fn=Force normal Ff=Force friction

**1. A 3500 kg car is travelling east at 90. km/h and realizes that the road is becoming
slippery so he puts his foot on the brake to come to a stop. He stops after 45 metres. Find the coefficent of friction between
the road and the tires on the road.**

**2. A 3250 kg car is driving at 50. km/h and the driver provides a force on the brakes
to bring the car to a stop at a stop sign. If the coefficent of friciton is 0.65 how far did the car go before stopping?**

**3. In a car testing lab the scientists want to find the force a seatbelt provides
on the crash test dummy. The car is travelling 90. km/h and the dummy weighs 80. kg. When the car stops it and the dummy travel
1.5 metres in the process. There is constant acceleration in the collision. Find the force the seatbelt exerts on the crash
test dummy.**

**4. A convertible model car runs into a telephone pole the passenger is not wearing
their seatbelt. As a result the passenger flys out of the car at a 20 degree angle at 5 m/s. Find the horizontal
distance the person will travel as they land on the side of the road. (Don't worry the person is fine!)**