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Homework Help: Acceleration: How do I know to use one time and not the other?

  1. Feb 13, 2012 #1
    There is a student running at a constant velocity after a stationary bus, when the student is a certain distance from the bus, the bus begins to accelerate. how many seconds does the student have before overtaking the bus.

    In order to solve the problem I have to combine acceleration formulas for the student and the bus.

    x + vt - (1/2)gt^2 = x + vt - (1/2)gt^2

    and solve for time. This ends up being a quadratic formula so I get two solutions for t.

    both solutions end up being positive.

    How do I know which one to use for the correct answer?
  2. jcsd
  3. Feb 13, 2012 #2


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    probably the earlier one. Think about it, when the bus is still at speed slower than the student, the student overtakes the bus. When the bus accelerates to speed faster than the student, the bus will overtake the student.
  4. Feb 13, 2012 #3


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    If you ran towards a bus which then accelerated, you will catch the bus, go past the bus, the bus will then pass you when it has got up to a higher speed and then it will move away.
    It was next to you twice, one when you caught it, and a second time when it caught you.
    That second case won't actually happen, since when you first caught the buss you probably jumped aboard. When we run to catch bus it is not usually just to show we can catch up, but so that we can climb aboard when we first catch it.
  5. Feb 14, 2012 #4
    Thanks wukunlin and thanks PeterO. lol, Peter, your response sparked a memory of that first superman movie where he's in high school, casually running/jogging beside a train lol. Thanks for the response, it makes perfect sense.

    I have another question that I wasn't even going to ask, but since the concept for this question made so much sense, i'll give another question a try.

    Can I ask a new question in this thread regarding the time variable in acceleration equations or should I start a new thread for it?
    Last edited: Feb 14, 2012
  6. Feb 14, 2012 #5


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    Try here
  7. Feb 14, 2012 #6
    Okay, here it goes:

    A rock falls from a cliff, the height of the cliff is unknown. The total time taken for the rock to fall the total distance is unknown. However, a person sees the rock as it's falling from the bottom 1/3 of the cliff to the ground and notes that it took 1.3 seconds for the rock to fall that unknown distance of 1/3 the total height of the cliff.

    So I don't know the total time or total height, but I know a time at a given fraction of total height.

    since 3/3 - 2/3 = 1/3, I let 3h/3 be total height and 2h/3 be the other height so that 3h/3 - 2h/3 = 1h/3

    Now I can set say 3h/3 = x + Vt - .5gt^2

    and 2h/3 = x + vt - .5gt^2

    If I solve both of those equations for t, I can say t= √(2h/g) and the other t = √(4h/3g)

    subtracting t - t = time of 1/3 total height in numbers, t - t = 1.3 seconds

    so √(2h/g) - √(4h/3g) = 1.3

    and solving for h gives total height of the cliff.

    However, I am confused about the time of total height and time of other 2/3 height. Both are square roots, so both have 2 solutions. So in total, there are 4 different solutions for time in this problem. I used two of the times to represent time fallen from total height and time it took to fall 2/3 of total height. But, what do the other two time solutions represent? - I'm not sure whether the other two time solutions are positive or negative. If negative, would I just disregard it for being a meaningless negative time? I could be wrong, but I think one of the unused negative square root time solutions, if used, would still produce a positive height value. What would that time and height value represent?
  8. Feb 14, 2012 #7


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    Suppose you were in a passing train and just saw this last 1/3 of the fall.

    You may assume it was something falling from the top of the cliff, but it may actually have been something fired, like a mortar, from the beach - which first went up to cliff top height, then cam back down again - and you just saw the last little bit.

    The same mathematical equations would be used to solve the question, and presumably you should get two times out on many occasions - one for on the way up [in case there was one] and one for on the way back down.

    Not sure if that applies here.

    I would be solving a little more simply I think
  9. Feb 14, 2012 #8
    From typing in the masteringphysics h/w problem into google this is the only description i can find of the what is occuring when you take the negative of the square root for one of the times. But it is not from the actual company who produced this cliff question, so I don't know if I can be one hundred percent sure of its correctness.

    But assuming it is correct, what does it mean. I don't understand the scenario that the words are describing.

    If somebody is throwing a rock up and it lands back on the ground at 1.3 seconds later, there must be some initial velocity that pushed it up into the air. But the original equation states that initial velocity is zero, so how can one assume a negative value for time means a rock was thrown upward, if they are basing that assumption on an equation that states initial velocity is zero?

    Last edited: Feb 14, 2012
  10. Feb 14, 2012 #9


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    Suppose you had a friend drop a stone from the top of a 10 storey building. Meanwhile you were waiting at a 7th floor window, and you had calculated what speed the dropped stone would pass you.
    Cleverly, you threw another stone down with exactly that speed and at just the right time, so the two stones accelerated to the ground side-by-side.
    A passer-by noticed the last 3 m of the stones' fall - but of course had no idea which stone began with zero velocity [dropped] and which one began its plunge being thrown.

    Indeed, had one of the stones actually got caught on an awning, the passer-by might just assume the stone that reached the ground was the one that fell - assuming no-one would be silly enough to throw a stone from a 7th storey window.

    That is why the formula for dealing with a rock dropping from a cliff, and one thrown up from the ground both give different, plausible answers.
  11. Feb 14, 2012 #10
    hi Peter, hope all is well. In this specific problem, the cliff is determined to be 246 meters. The bottom third would then be a height of 82 meters.

    I completely agree that if someone were at the window on the seventh story of a 10 story building and threw a stone at the same time as a dropped stone reached the 7th story, then both stones would reach the floor at the same time, and someone passing by would not know which stone was thrown and which one was dropped - hopefully the passerby is wearing a hard hat.

    However, the bottom third of this cliff is 82 meters, so if someone was at an 82 meter mark on the cliff and threw a stone with the same velocity, then both stones would take 1.3 seconds to hit the ground. Completely makes sense. However, 82m is definately much higher than 2.51m. Surely you aren't implying that he threw the stone at the same velocity starting from 2.51 meters, it would only take a small fraction of the time for the stone to cover the distance of the bottom 2.51 meters of the cliff. And I definately would probably never be able to throw fast enough to perfectly match the velocity :)

    So, where does the height of 2.51 meters come into the picture?
    Last edited: Feb 14, 2012
  12. Feb 14, 2012 #11


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    Where are you getting this 2.51 from?

    EDIT: I didn't read that bit of typing carefully. What question did that relate to?
    Last edited: Feb 14, 2012
  13. Feb 14, 2012 #12


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    IS that your answer, or an answer from the text?
  14. Feb 14, 2012 #13
    Both of course :)

    It is in fact an answer that corresponds with the text.

    246 meters is the height of the cliff.

    Within the problem there are two different times to solve for. Time it takes the rock to fall the total distance. And Time it takes the rock to fall 2/3 the distance.

    The difference between those two times is equal to the time it takes for the rock to fall the final 1/3 distance.

    So setting the equations for (time of total distance) - (time of 2/3 distance) = (time of final 1/3 distance)

    Then, solving for height produces 246m.

    Then there is a part b to the question, which asks something like:

    When solving for 2/3 time, you get a positive and negative time. If you use the negative time instead of the positive time to find height, what does that height represent?

    And the text I attached, is the supposed answer I found on the internet.

    So instead of (time of total height) - (negative time of 2/3 height) = (time of final 1/3 height) And then solving for height produces 2.51 meters.

    And the question askes what that height is referring to.
  15. Feb 14, 2012 #14


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    OK I get it now - I just hadn't seen you derive that final figure of 246 m, which I had calculated but not shown you.

    Note: Did you also see the g value cancel out as you worked through the problem, meaning that if the final 1/3 of the drop is covered in 1.3 seconds, then the full drop lasted 7.08434 seconds regarless of which planet you conducted the experiment.
    Of course in weaker gravity on the moon, the cliff would not have been as high, and in stronger gravity on Jupiter - it is even higher.
    I will now try to find the 2.51 for you.
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