A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together and move 1.97 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

(1)

Where:

, - Masses of the cars, in kilograms.

, - Initial velocities of the cars, in meters per second.

- Velocity of the resulting system, in meters per second.

If we know that , , and , then the  velocity of the resulting system:

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy (), in joules, is dissipated due to work done by friction (), in joules, that is to say:

(2)

(2b)

Where:

- Coefficient of friction, no unit.

- Gravitational acceleration, in meters per square second.

- Travelled distance, in meters.

If we know that , and , then the coefficient of friction is:

The coefficient of friction between the cars and the road is 0.859.

e) none of the above answers are correct

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min