Physics College

A 940-kg sports car collides into the rear end of a 2500-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
What was the speed sports car at impact?
Express your answer to two significant figures and include the appropriate units.

Answers

Answer 1

The speed of the sport car at the time of impact is 6.61 m/s.

What is the frictional force of the two cars?

Frictional force of the two cars = coefficient of kinetic frictin × mass × acceleration of gravity

= 0.8 × (2500+940) × 9.8

= 26970N

What is the acceleration of the skidded cars?As per Newton's second law of motion, force = mass × accelerationAcceleration= force / mass

= 26,970/3440

= 7.8 m/s²

What is the velocity of the sport car at the time of impact?As per Newton's equation of motion, V² - U² = 2aSHere, V = 0 m/s, a= -7.8 m/s², S= 2.8 mSo, 0²-U²= 2×(-7.8)×2.8

=> U = √43.68

= 6.61 m/s

Thus, we can conclude that the speed of the sport car at the time of impact is 6.61 m/s.

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Related Questions

A block and tackle pulley system has 3pulley wheels in the lower movable block. Determine the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​

Answers

A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort

How to calculate  the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​?

VR =3

VR = n=3

Efficiency of the system = 80%

Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4

Man can lift load with effort =350N

Thus,

Load= MA× effort = 2.4×350 = 840 N.

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