Question

We will be exploring the following road/weather/tire conditions:

1. Dry Road / New Tires
2. Dry Road / Old Tires
3. Raining / New Tires
4. Ice / Average Tires

We will be keeping the mass of the car fixed at 1000kg and changing the initial speed of the car. Our goal is to see how the braking or stopping distance of the car depends on the initial speed and on the road/weather/tire conditions.

a) Rank the road/weather/tire conditions in the order of smallest to largest stopping distance. Note that this is just a qualitative prediction. It is okay if it later turns out to be wrong. We want to make a hypothesis based on our everyday experience and see if we turn out to be right! It is especially interesting to think about whether rain or old tires will have more of an effect in increasing the braking distance. Write the order you think the braking distances will be in below.

b) Derive a relationship/formula for the dependence of force of friction of the road surface on the mass, initial speed and displacement of the car using each of the two methods below. Note that this is exactly the same as problem 3 in ‘Worksheet – Energy Conservation, Work-KE Theorem’. The only difference is now you are solving for force of friction instead of initial velocity.

1) derive using dynamics-kinematics method

2) derive using work-kinetic energy method

The equation you get from the two methods should be the same.

Answer 1

a) As tires become old, the surface becomes more flat, so surface area in old tires is greater than the new ones, it offers more friction in dry roads. In rainy roads the water on roads provide lesser friction than dry roads. And in ice it is even lesser than rainy roads. Because at the contact point of tires an the ice, the I've melts due to the pressure and the situation becomes as if the car is running on water only. So in the order of smallest to largest stopping distance the list will be -

i) Dry Road / Old Tires

ii) Dry Road / New Tires

iii) Raining / New Tires

iv) Ice / Average Tires

b) 1. Let the coefficient of friction between the road and the tires is μ, the mass of the car is m, initial velocity of the car is u, the stopping distance is so and acceleration due to gravity is g. So normal force on the car by ground is N=mg

Then after applying the brakes the frictional force on the car is μ.N=μmg. So the deceleration of the car will be (μmg)/m=μg. As the final velocity of the car will be zero, the equation of motion of the car will be -

  $0=u^{2}-2\mu gs$

$\boldsymbol{\Rightarrow u^{2}=2\mu gs}$

2. Now, initial kinetic energy of the car was $\frac{1}{2}mu^{2}$ . The car has gone a distance s against the frictional force 'μmg'. So the energy dissipated in this process is 'μmgs' and as the final kinetic energy of the car is zero, the dissipated energy will be same as the initial energy, that is -

$\frac{1}{2}mu^{2}=\mu mgs$

  $\boldsymbol{\Rightarrow u^{2}=2\mu gs}$