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Simple velocity question

  1. Jun 8, 2008 #1
    Say I wanted to know how fast I would need to go in a car that weighs 6000lbs that would clear at 20 foot gap in a bridge inclined 10 degrees. Is there a way I can use conservation of energy to find the velocity needed to clear the 20 foot gap?


    Would I put the 20ft in the h spot? Since I do not know the acceleration of the car I cannot use kinamatics. Any help would be appreciated.

    Thank you.

  2. jcsd
  3. Jun 8, 2008 #2


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    Hi StephenDoty,

    No, 20ft would not be in the h spot; h is the vertical height.

    Does this mean it's like a drawbridge that's partially open (so it's 10 degrees upwards before the gap and 10 degrees downwards after the gap)? I think kinematics would be the best way to go. Once the car has begun the jump you do know it's acceleration; unless I'm misreading your question, all you are looking for is the velocity right before the jump.
  4. Jun 8, 2008 #3
    right I am looking for the velocity right before the jump that would allow the car to land on the other side. The problem shows a regular inclined bridge that got damaged and now there is a 20ft hole between the two sides. The two sides are raised upward from the horizontal 10 degrees. Since I do not know the acceleration of the car from when it starts going up the first part of the bidge until the car starts the jump, at which time a=g. How would I find the velocity needed to clear the gap?

    Thanks for the replies
  5. Jun 8, 2008 #4


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    I may be visualizing your problem wrong; let me know if I am.

    We don't need to know the acceleration of the car before the jump; let's just say there was no acceleration along the road and that it was going at a constant speed the entire time before the jump, and that speed is what is being asked for.

    Just if I'm visualizing this correctly, isn't [itex]\Delta y=0[/itex] for the jump itself (the landing point is the same height as the initial point)? If that's right, then you have [itex]\Delta x[/itex] and the angle of the initial velocity, and can find the speed.
    Last edited: Jun 8, 2008
  6. Jun 8, 2008 #5
    Hi Stephen,
    At the point you land on the other side, the vertical component of your velocity should become zero due to gravitational retardation (9.81 m/sec^2), if you ant to get the minimum value of velocity.
    v sin 10 - 9.81*t=0 ......... (i)
    v cos 10*t=6.096 m (=20 ft) ........ (ii)
    t= 6.096/(v cos 10) {From eq. (ii)}
    Put this value of t in equation (i) and solve for v

    And by the way dont try it practically :wink:
  7. Jun 8, 2008 #6
    This was when the following diagram holds:

    Attached Files:

    • C1.JPG
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  8. Jun 8, 2008 #7
    If the second case holds: (See Diagram)
    6.096 / (v cos 10) =t
    6.096 tan 10 = (v sin 10) * t + 0.5 * 9.81* t*t

    Solve the quadratic for t and substitue the value in the previos equation to get minimum "t"

    6.096 metres= 20 feet
    therefore the answer u will get will be in metres per second which u can convert back to feet per second.

    Attached Files:

    • C2.JPG
      File size:
      3.1 KB
  9. Jun 8, 2008 #8


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    Hi ritwik06,

    I don't think there is a solution for the second case you talked about. As soon as the car reaches the gap, it's vertical velocity will start decreasing, and so the trajectory will miss the other side. (In your equation, you need a minus sign:

    6.096 tan 10 = (v sin 10) * t - 0.5 * 9.81* t*t

    but the left side won't happen because the car can never make it there.)

    I had it visualized differently:


    which is really exactly the same as your first case. (But we don't need the vertical velocity to be zero at the end of the jump. If it's at the same height at the end, the speed in the vertical direction must be the same as at the beginning. It will just be in the opposite direction.)


    Could you post a diagram?
  10. Jun 8, 2008 #9
    Oh yes, I am sorry for the sign mistake. it is indeed negative. I dont think so that the trajectory will miss the second side. Indeed the vertical component of the velocity will start decreasing but if the component sufficiently large to travel the vertical distance, the trajectory wont miss the target.

    Yes, again I admit my mistake in the second case. You are very right. The velocity would be the same as before but in negative direction.

    the equations would be:

    20/ (u cos 10 ) = t
    0=u sin 10 *t - 0.5*9.81*t*t

    I think it is correct now.

    I am very sorry for the mistakes. I would be careful from now on.
  11. Jun 8, 2008 #10


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    If you don't mind discussing this a bit further: I think it will never be able to cross the gap in the case that you have in post #7. Here's my reasoning: Imagine there is no gap in the bridge. Then the car travels in a straight line.

    Now with the gap, the car follows a parabolic trajectory, that is always below that straight line. So the instant it is in the gap, the car begins falling away from that straight line and cannot make it across. What do you think?

    That's exactly what I got. Hopefully Stephen can post a picture or verify that our diagrams are what the problem was talking about!
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