Coefficient of friction and drunk driver

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SUMMARY

The discussion centers on calculating the speed of a drunk driver just before a collision using the equation Mv²/2 = (M+m)μg * S. The user correctly identified the masses of the vehicles as 2400 kg and 1200 kg, the coefficient of friction as 0.70, and the skid distance as 25 m. However, the user initially neglected to include the acceleration due to gravity (g = 9.81 m/s²) in their calculations, leading to an incorrect speed of 7.246 m/s. The correct application of the equation, including gravity, yields a different result.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of kinetic friction
  • Knowledge of basic algebra and equation manipulation
  • Ability to apply physics equations in real-world scenarios
NEXT STEPS
  • Review the principles of conservation of momentum in collisions
  • Learn about the role of friction in motion and stopping distances
  • Study the effects of mass and velocity on collision outcomes
  • Explore the implications of different coefficients of friction in various materials
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as professionals in automotive safety and accident reconstruction.

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Homework Statement



A drunk driver in a 2400kg car plows into a 1200kg parked car with its brake set. Police measurements show that the two cars skid together a distance of 25m before stopping.

If the effective coefficient of friction is 0.70, how fast was the drunk going just before the collision?


Homework Equations



Mv^2 / 2 = (M+m)μg * S i used this equation

2400*v^2/2=(2400+1200)*.70*25m

and i got 7.246 for v

what did i do wrong?

]
 
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It looks to me like when you subbed in your values for the equation, you forgot acceleration due to gravity.
 

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