Iran: Passenger Crushed By Drifting Car

superhamster · #51 ● 11-23-2021, 05:03 PM

That was HILARIOUS!

Umea · #52 ● 11-23-2021, 09:54 PM

Originally Posted by DanteAntonelli ↗

The long answer would be:

The friction force that is required to stop that clunker would be equal to its kinetic energy
Let's say the car weighs 2,000 Kg
The kinetic energy of the car is
KE= 1/2 m v2
where:
KE = Kinetic energy of the drifting car
m = mass (kg)
v = velocity (m/s)

KE = 1/2 (2000 kg) ((100 km/h) (1000 m/km) / (3600 s/h))2
= 771605 J

The friction energy needed to stop the car can be expressed as
FE=FF SD
Where
FE= Friction energy to stop the car (Joules)
FF= Friction force (Newtons)
SD= Stopping distance (m)

Since the kinetic energy of the car is transformed into friction energy, we can say:
KE = FE
To calculate the friction force FF
FF = μ m g
Where:
FF= Friction force
μ - Friction coefficient (0.6 for dry concrete)
m = Mass of object
ag = Acceleration of gravity

= 0.6 (2000 kg) (9.81 m/s2)
= 3924 N
So the distance would be
d=FE/Ff
d=771605 J/3924 N = 197m

Now considering the following kinetic friction coefficients for rubber on dry and wet concrete
On dry concrete & rubber μ=0.6 - 0.85
11,772 N - 16677 N friction force
On wet concrete & rubber μ=0.45 - 0.75
8829 N - 14715 N friction force

We can conclude that the needed distance would be longer for lower friction coefficients, we can confidently assume that the body under the car is turned to a mushy wet goo, hence the car would be sliding an a wet mushy substance and that only means a longer distance would be needed for the clunker to get to a stop.