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Testifying in a case involving an accident

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  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data

    As an expert witness, you're testifying in a case involving a motorcycle accident. A motorcyclist driving in a 60km/h zone hit a stopped car on a level road. The motorcyclist was thrown from his bike and landed 39m down the road.
    Was he speeding?


    3. The attempt at a solution

    Capture.JPG
     
  2. jcsd
  3. Jan 28, 2014 #2
    Can you elaborate on your work and what exactly you're getting hung up on?
     
  4. Jan 28, 2014 #3
    I have the range and the expression for the range. If my work is correct, as attached, until hitherto, I'm having problem finding vi or Θ.
     
  5. Jan 28, 2014 #4
    Based on the structure of the question I would assume that, they don't intend for you actually find the exact speed they were travelling. So I would just start by assuming they were going 60km/h and see if that would cause them to travel 39m.
     
  6. Jan 28, 2014 #5
    I suppose that would be a reasonable premise from which I can work on. It should be solvable then.
    Will post solution in the morning.
     
  7. Jan 28, 2014 #6
    That's strange. I was only able to deduce the angle, assuming the entity was travelling at 60kmh^-1 and landed 39m from point of collision, at which the entity was flung upon collision.
     
  8. Jan 28, 2014 #7

    Curious3141

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    Homework Helper

    You don't know the angle of launch (θ).

    What's the formula for range? How would you find the maximal range from that formula (hint: max value of sine)?

    Now see what the max range for a vi of 60km/h is.
     
  9. Jan 28, 2014 #8
    I didn't notice the first half of the attachment were work from another question.

    Anyway,
    tfull trajectory = [itex]\frac{2 vi sin Θ}{g}[/itex]

    x(tfull trajectory) =
    [itex]\frac{2vi^2 sinΘcosΘ}{g}[/itex] = [itex]\frac{vi^2 sin(2Θ)}{g}[/itex]

    I know [itex]39m [/itex] = [itex]\frac{2vi^2 sinΘcosΘ}{g}[/itex] = [itex]\frac{vi^2 sin(2Θ)}{g}[/itex]

    In simplying, [itex]382.2m[/itex] = [itex]vi^2 sin(2Θ) [/itex]

    I'm not entirely sure it's right to substitute [itex] (60kmh^-1 = 16.67ms^-1) [/itex] into the above equation. If I did, then [itex] sin(2Θ) = 1.38 [/itex] which has an obvious issue.

    Edit: Sine 45? It did occurred to me but in theory, would it be reasonable to assume the entity is flung at an angle of 45 degrees?
     
    Last edited: Jan 28, 2014
  10. Jan 28, 2014 #9
    Another hypothesis:

    If I assume launched angle is 45° and initial velocity = [itex] 16.67ms^-1[/itex]

    and

    x = [itex]\frac{2vi^2 cosΘsinΘ}{g}[/itex]

    then [itex]16.68ms^-1 cos 45° \frac{16.68 sin 2(45°)}{9.8ms^-2}[/itex]

    = 28m

    Given the above parameter, the displacement is 28m

    Therefore, given the same launch angle of 45°, the only possible solution for the entity to be flung at a displacement > 28m is for the initial velocity to be [itex] > 16.67ms^-1 [/itex]

    Hence, it can be deduced that the entity was speeding.

    and given the entity achieved a displacement of 39m and suppose we further assume the launch angle is 45 degrees, the initial velocity works out to be 19.6ms^-1 and 1.96ms^-1 > 16.67ms^-1.
     
    Last edited: Jan 28, 2014
  11. Jan 28, 2014 #10

    Curious3141

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    If θ = 45 degrees, what is 2θ and therefore sin 2θ in the formula? Is this the maximum value of the sine function?

    You're not saying he was definitely flung at that angle. You're saying that this (45 degrees) is the angle that maximises range at any initial velocity. If he was going at the limit, even this maximal range would fall short of how far he was actually propelled. Hence, what can you conclude?
     
  12. Jan 28, 2014 #11
    I did further work 2 posts earlier
     
  13. Jan 28, 2014 #12

    Curious3141

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    This post is basically right, except for a couple of typos. You computed the minimum velocity that he would actually have had to be travelling at in order to achieve that range. Not strictly necessary, but good to know. Note that he could well have been travelling faster than this (in which case his launch angle might have been lower or higher than 45 degrees).
     
  14. Mar 24, 2014 #13
    hey guys ive got the same question but the first part of the question ask for "Find the minimum speed he could have been going just before the accident." can someone please help?
     
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