The Shortest Stopping Distance

Classical Mechanics Level 1

Consider a car moving along a straight horizontal road with a speed of 72 km / h 72 \, \text{km}/\text{h} . If the coefficient of kinetic friction between the road and tires is 0.5 0.5 , find the shortest distance over which the car can be stopped in meters.

Take g = 10 m / s 2 . g = 10\ \text{m}/\text{s}^2.


The answer is 40.

Abhiram Rao by Abhiram Rao , 18, India

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2 solutions

Alex Boone
Jan 29, 2019

v=72km/h Coefficient of kinetic friction=0.5 g=10m/s d=(v×v)/(2×g×Coefficient of friction)

v=72km/h÷3.6=20m/s

d=(20×20)/(2×0.5×10)=400/10=40m

d=40m

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Sahil Singh
Jul 19, 2018

velocity of car = 72km/h = 20 m/s... we know that any force = ma , in this case decelaration would be there due to fricion : friction = uN... m a=u N... m a=u m*10 ( as N= mg and taking g=10m/s^2)... so, a=5m/s^2 (as u=0.5)... we get v^2=u^2 +2as... > 0=400 + 2(-5)s... so, s=40m ...

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