Testifying in a case involving an accident

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SUMMARY

The discussion revolves around calculating whether a motorcyclist was speeding at the time of an accident where he was thrown 39 meters after colliding with a stopped car in a 60 km/h zone. Participants utilized physics formulas for projectile motion, specifically the range formula, to deduce that if the motorcyclist was launched at a 45-degree angle, the minimum initial velocity required to achieve a 39-meter displacement would be approximately 19.6 m/s. This indicates that the motorcyclist was indeed speeding, as 19.6 m/s exceeds the converted speed limit of 16.67 m/s (60 km/h).

PREREQUISITES
  • Understanding of projectile motion and its equations
  • Familiarity with the range formula: x = (vi^2 * sin(2Θ)) / g
  • Knowledge of basic trigonometry, particularly sine and cosine functions
  • Ability to convert units, specifically from km/h to m/s
NEXT STEPS
  • Research the physics of projectile motion and its applications in accident reconstruction
  • Learn about the implications of initial velocity and launch angles in real-world scenarios
  • Study the conversion of speed units, particularly from kilometers per hour to meters per second
  • Explore case studies involving expert witnesses in accident analysis
USEFUL FOR

This discussion is beneficial for physics students, accident reconstruction experts, legal professionals involved in personal injury cases, and anyone interested in understanding the dynamics of motorcycle accidents.

negation
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Homework Statement



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?


The Attempt at a Solution



Capture.JPG
 
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Can you elaborate on your work and what exactly you're getting hung up on?
 
MostlyHarmless said:
Can you elaborate on your work and what exactly you're getting hung up on?

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.
 
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.
 
MostlyHarmless said:
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.

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.
 
MostlyHarmless said:
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.

That's strange. I was only able to deduce the angle, assuming the entity was traveling at 60kmh^-1 and landed 39m from point of collision, at which the entity was flung upon collision.
 
negation said:

Homework Statement



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?


The Attempt at a Solution



View attachment 66096

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.
 
Curious3141 said:
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.

I didn't notice the first half of the attachment were work from another question.

Anyway,
tfull trajectory = \frac{2 vi sin Θ}{g}

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

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

In simplying, 382.2m = vi^2 sin(2Θ)

I'm not entirely sure it's right to substitute (60kmh^-1 = 16.67ms^-1) into the above equation. If I did, then sin(2Θ) = 1.38 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:
Another hypothesis:

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

and

x = \frac{2vi^2 cosΘsinΘ}{g}

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

= 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 > 16.67ms^-1

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:
  • #10
negation said:
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?

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?
 
  • #11
Curious3141 said:
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?

I did further work 2 posts earlier
 
  • #12
negation said:
Another hypothesis:

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

and

x = \frac{2vi^2 cosΘsinΘ}{g}

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

= 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 > 16.67ms^-1

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.

This post is basically right, except for a couple of typos. You computed the minimum velocity that he would actually have had to be traveling at in order to achieve that range. Not strictly necessary, but good to know. Note that he could well have been traveling faster than this (in which case his launch angle might have been lower or higher than 45 degrees).
 
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  • #13
hey guys I've 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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