How to Calculate a Car's Speed from Skid Marks on Wet Pavement

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Homework Help Overview

The discussion revolves around calculating a car's speed based on skid marks on wet pavement, specifically focusing on concepts of kinetic energy and energy conservation. Participants are exploring the physics involved in the transition from motion to rest and the role of friction in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of kinetic energy formulas and the relationship between speed, mass, and work done by friction. Questions arise regarding the correct formulas to use and the assumptions about friction coefficients.

Discussion Status

Some participants have provided hints and guidance on using energy conservation principles to relate kinetic energy to work done by friction. There is an ongoing exploration of the necessary information, such as the friction coefficient, to solve the problem effectively. Multiple interpretations of the problem setup are being considered.

Contextual Notes

Participants note the absence of specific information, such as the coefficient of friction, which is crucial for solving the problem. There is also mention of the original poster's challenges in catching up with missed material, which may affect their understanding of the concepts discussed.

Moe_slow
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i have a phyiscs test soon( in high school Grade 11) and am stuck on this one question:

*A 1500kg car accelerates uniformaly from 0 to 100km/h in 6.0s*
a) how much energy does the car have after accelerating?

i try to follow the formula F=ma but i am not getting anywhere...i convert 100 km/h to m/s and then multiply by 1500kg...but am ending up with a formula diff answer than the one shown...am i using the wrong formula?... help would be apperciated...
 
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HINT: How fast is the car going after it is accelerated and what is its kinetic energy when it is going that fast? :)
 
Yes, you are using the wrong formula

[tex]p = mv[/tex]

That is the formula you used, and it is for linear momentum, not kinetic energy.

This is the formula for kinetic energy:

[tex]KE = \frac{1}{2}mv^2[/tex]
 
thanks i got it...

i did:

KE= 1/2 mv^2
= (1500kg)(27.7m/s^2)/2
= 578703 > 5.8 x 10^5

thanks a lot...i think i became brain dead after all the work...gota stop using wrong formula...
 
The kinetic energy has nothing to do with the amount of time that it took to get to a certain speed (to the acceleration of the mass) , only to the speed at a given time.
 
could i get some more help to get me started on this question...i missed quite a bit of time at school and now am trying to catch up...some help would be appreciated to get me back on track for the test...

here is the question:

*a car 925 kg without anti-lock brakes has skid marks 15.3 meters long on wet pavement.*
a) wat was the cars speed at the start of the skid.

i need some kind of a direction to start this off with...i am a bit of thrown back with this one
 
To solve that question you first need to know the friction coefficiant of the tires on the cement. With that you can find the friction force
(F= SC*N) and then the Work that the pavement did on the car (W=Fx)
Since the car stopped in the end you know that all of its original kinetic energy was conserved to heat energy by the work done by the pavement so you know that the original energy: Ek = 0.5mv^2 equals the total work: W=F*x . With the Ek and mass you can find the speed.
*I'm assuming that they want you to solve this with energy conservation.
 
Moe_slow said:
could i get some more help to get me started on this question...i missed quite a bit of time at school and now am trying to catch up...some help would be appreciated to get me back on track for the test...

here is the question:

*a car 925 kg without anti-lock brakes has skid marks 15.3 meters long on wet pavement.*
a) wat was the cars speed at the start of the skid.

i need some kind of a direction to start this off with...i am a bit of thrown back with this one
Use energy conservation first-- kinetic energy will all be converted to work, thus:
KE - W = 0
Or KE =W
KE = 1/2 mv^2, W = m|a| * x
1/2 mv^2 = m|a| x
Cross out the m s and do a little manipulation:
v = root(2|a|x)
You would have a and x to solve the problem, but apparently you aren't given a according to your question? (If you were given the co-efficient of friction as the above poster said, you would have a = mu * g, where mu is the co-efficient of friction, and thus you can solve the problem)
 

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