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Indicating negative direction on a velocity-time graph

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball is released from rest and falls freely to the floor. It then rebouces to the same height. The graph used for representing the above is saved as an attachment (1). With that graph in mind, how do I draw the general velocity-time graph of a ball that moves at constant velocity along a horizontal track. It then moves along an downward inclined plane, bofore rolling onto another horizontal track , an upward inclined plane and finally another horizontal track. Take all the surfaces to be frictionless and connected (I have drawwn the path travlled by the ball also (2)).

    3. The attempt at a solution
    I've attached two attemtped graphs that I came up with but I'm not sure which is correct. Pls tell me which one is right and why.
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2008 #2

    Mentallic

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    For the velocity-time graph, just try and imagine how the ball would move in this scenario, also using your knowledge of acceleration due to gravity and how this is represented on this type of graph. At the instant the ball is dropped from a height h, what is its velocity? how does the velocity change as it begins falling to the ground? What happens in the split second when the ball hits the ground and bounces back up? how does the ball move while on its trip back up, and what happens at the instant the ball is back at height h?
    You have to keep in mind that velocity is a vector (it has direction) so you will need to assign a direction (positive or negative) to the downward fall and then the opposing direction to the upward bounce.

    As for the planes, just think about how gravity affects the ball in these situations. Obviously going uphill will slow the ball down.
     
  4. Oct 11, 2008 #3
    Thank you for your response. However, I am not sure why, in the example graph, there is a sharp horizontal straight line downwards. How does the acceleration decrease to zero and then increase in the opposite direction? If the graph is true, attempt 2 for the second part of my previous post should be correct since the ball changes its direction (to negative) and then increases from there. Am I right?
     
  5. Oct 11, 2008 #4

    Mentallic

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    This sharp change in velocity in the example graph is the split-second the ball bounces. Its velocity is at a maximum in the negative direction (downward towards the ground) and quickly slows down and bounces back in the opposite direction (upwards) at full speed. As you can see, the velocity passes through the 0m/s on the bounce, so you can conclude that for the ball to change its direction, it needed to at some instant be stationery.

    For your actual problem, it is quite easy to choose between those 2 graphs you proposed, simply because the ball never turns to change its direction :wink:
     
  6. Oct 11, 2008 #5
    Thanks for the response again, but doesn't the ball change direction when it moves up the plane (it moves down, along a straight, horizontal track and upward along another plane). I know that the velocity has to decrease as it moves up the second plane, but why doesn't isn't the ball moving in negative direction?
     
  7. Oct 11, 2008 #6

    Mentallic

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    Oh yes that part didn't cross my mind. The simple answer is: the velocity vector is only considering the horizontal motion, not the vertical. Vectors do include angular motion, but that is a little too complicated and unnecessary for this question.

    There is also a reason why you can assume only the horizontal velocity is being graphed. Take a look at the 2 graphs again. Both begin with a constant positive velocity, but the ball is travelling on a flat, horizontal plane. The vertical displacement is 0, so therefore the velocity being graphed is horizontal, or east-west.

    or maybe we can take the advice from my 11 year old brother? "#1 because that looks like that other one, but just flipped around"

    EDIT: after a bit more thinking, the vertical direction could be playing a role. The constant gradient on the graphs suggest uniform acceleration, but this isn't possible if just taking the horizontal velocity into account. Now I'm confused too...
     
  8. Oct 11, 2008 #7

    Redbelly98

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    If the upward slope continued indefinately, eventually the ball would reverse direction and roll back down.

    However, the problem statement says the ball reaches the final horizontal track. So it must make it all the way up the slope without reversing direction.
     
  9. Oct 11, 2008 #8
    Thanks for responding but I don't get what you mean. The ball cld have just decreased it's velocity to 0m/s in the negative direction and then increase it's velocity in the positive direction if it were to eventually reverse direction and roll back down on the upward plane. Pls explain. Thank you very much in advance.
     
  10. Oct 12, 2008 #9

    Mentallic

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    If you consider the ball travelling towards the east (left to right) as being a negative direction, then the graph could not be showing a positive velocity at the start.
    I hope this is what you were asking.
     
  11. Oct 12, 2008 #10
    I've finally got it! Thank you! Thank you! Thank you!
     
    Last edited: Oct 12, 2008
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