Calculate Car's Speed in Collision: Brakes and Skid Marks Analysis

  • Thread starter Stroodle
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In summary, to find the initial velocity of a car that hit a truck after slamming on its brakes, we can use the kinematic formula v^2=u^2+2ax, where u is the initial velocity, v is the final velocity, a is the constant acceleration, and x is the distance traveled. Plugging in the given values of 30\ ms^{-1} for u, 40\ m for x, and -6\ ms^{-2} for a, we get a final velocity of approximately 20.5 \tfrac{m}{s}. Remember to include the exact answer along with the approximation.
  • #1
Stroodle
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A driver traveling at [tex]30\ ms^{-1}[/tex] slammed on the brakes when a truck pulled out ahead. The skid marks leading to the wreckage are [tex]40\ m[/tex] long. Tests on a similar model car show the brakes provide a steady deceleration of [tex]6\ ms^{-2}[/tex]. How fast was the car traveling when it hit the truck?

I'm not sure how to go about solving this problem, so any help will be greatly appreciated.

Thanks
 
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  • #2
What kinematic formula do you know tying in initial velocity (Given in the question), final velocity (This is your unknown), distance traveled (Given in the question) and the constant acceleration under which it was traveled (Also a given)?

Once you remember what this equation is, the problem will become as simple as plugging in the numbers.
Look over your class notes, it should be in there somewhere, and to give you another hint, it involves the squares of the velocities.
 
  • #3
Ah. Ok thanks for that.

So using the formula [tex]v^2=u^2+2ax[/tex]

I have [tex]v^2=30^2+2(-6)(40)[/tex]
[tex]\therefore v\approx 20\ ms^{-1}[/tex]

Is that correct?

Thanks
 
  • #4
Yes, that is correct. :)
[tex]V_f=\sqrt{420}\approx 20.493\approx 20.5 \tfrac{m}{s}[/tex]

Just make sure that if it's on a test, or on an assignment to be graded, that you include the exact answer along with the approximation.

You're very welcome. :)
 
Last edited:
  • #5
Will do. Thanks so much for your help.
 

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