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Get distance travelled after 3 seconds and velocity after 25m travelled

  1. Jul 23, 2008 #1
    1. The problem statement, all variables and given/known data
    If a stone is thrown vertically upward from the surface of the moon with a velocity of 10m/s, its height (in meters) after t seconds is h = 10t - 0.83t^2.

    a) What is the velocity of the stone after 3s?
    b) What is the velocity of the stone after it has risen 25m?

    2. Relevant equations
    h = 10t - 0.83t^2

    3. The attempt at a solution
    a) was easy and I got that. I took the derivative of h and used that velocity function (assuming h is the position function), to get v(3) = 5.02 m/s, which is correct according to my book.

    b) is giving me a load of trouble though. I've tried all kinds of equations, like the following:

    h = 10t - 083t^2 = 25 and solving for t

    v(t) = 10 - 1.66t = 25 and solving for t

    and a few others that don't make sense, like plugging 25 into the function for t, but since this is a function of distance depending on time elapsed, the input variable expected by the function is a value of seconds. In other words, I understand I can't just plug in a distance of 25m into a function that uses an integer of seconds to give back distance travelled.

    The answer in the back of the book is [tex]\sqrt{17}[/tex] which is equal to about 4.123...

    Nothing I've tried so far has given me back that number.

    Now, since the original position function h has as its output a distance of meters, it would make sense to solve for t where h = 25, and then taking the result of that, call it x, and solving for t with v(t) = x, no? I'm pretty sure I've tried that and it still hasn't worked out.

    What am I missing?
  2. jcsd
  3. Jul 23, 2008 #2


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    Blue: Yes this is valid to do. Why did you think that you had to put h(t)=25? What values did you get?
    Red: 25 is a distance, v(t) is a velocity at a time t, so you can't put a distance equal to a velocity and expect to get anything

    For your equation to be consistent, the units on the left must be the same on the right.
    For 10t, for this to be a distance, the 10 must have units of ms[itex]^{-1}[/itex] and the 0.83 to be in ms[itex]^{-2}[/itex].

    When you solve h(t)=25, you'll get values of t (what do these values represent?)
    When you answer that, you should be able to get it out.
  4. Jul 23, 2008 #3
    Hello again, and thank you for your continued effort to help! :). I appreciate it.

    When solving for h(t) = 25, I get two values t.

    t = 3.54029781... or t = 8.5078949...

    They represent the time t it takes for the object thrown to reach a height of 25m, right? Now, how can there be two values for this? How can an object thrown travel the same distance after two different amounts of time?

    So I'm trying to imagine this.. I'm standing on the moon and I throw a rock straight up. I wait about 3.5 seconds and then measure how far the stone has gotten since I threw it. I then wait until 8.5 seconds have passed, measure the distance traveled again, and to my surprise find that the distance remains the same as what it was on my first measurement! In other words.. once the stone has spent 3.5 seconds in the air, it comes to an abrupt stop and does not go anywhere until at the very least 8.5 seconds have passed.. what happens after that I do not know.

  5. Jul 23, 2008 #4


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    Let's not be on the moon for this one, let's be on Earth :smile:

    If take that same rock and throw it up in the air, it doesn't just keep going up right? It reaches a maximum height and falls back down. The motion is parabolic. Meaning that the equation show the upward motion for the rock reaching the maximum height as well as the downward motion for the rock. Do you see how at the two different times, the rock can be 25m from ground?
  6. Jul 23, 2008 #5
    Haha.. ok. Let's be on Earth :)

    Yes, I absolutely see that. But that means that the object would spend about 5 seconds just hovering in the same spot! That's just strange when imagined.. when I imagine that maximum point of an object's path of motion when thrown, my admittedly unscientific estimate of the time the object would spend at its maximum is just a small fraction of 1 second. But 5 seconds?
  7. Jul 23, 2008 #6


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    It doesn't.

    at t=3.5 seconds, the rock is still travelling upwards i.e. it hasn't reached the maximum height yet. Only after it reaches the max height, does it direction of travel change i.e. it starts falling back down where at t=8.5 seconds, it is 25m away from the ground again.
  8. Jul 23, 2008 #7
    Arg.. of course! I was on the phone with a friend when I mentioned to her this problem.. and she ended up explaining it in the exact same way. What I didn't realize is that the maximum could, and in fact is, in this case, higher than 25m. When you first mentioned the maximum, I just assumed you'd meant the 25m distance itself.. and of course you see that when you think about it like that, the two values for t make no sense.

    This makes perfect sense now! And since the question asks for the velocity after the stone has RISEN 25m, I assume I simply have to use the first value of t, which is about 3.5 seconds.

    Confirmed! Plugging 3.5 into v(t) gives me 4.19. sqrt(17) is 4.1231056..., so it's not exact, but it's the right idea.

    Now.. I have a TI-NSpire CAS calculator.. when I use it to solve h = 25, it gives as a result both values for t at once, separated by "or".

    What I'd like to know is.. how do I arrive at the exact result of sqrt(17)?

    When I do the calculation on paper, I end up at

    t(10 - 0.83t) = 25 for h = 25

    and don't see how to arrive at a result of sqrt(17) from there.

    Thanks again for the explanation!
  9. Jul 23, 2008 #8


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    t(10 - 0.83t) = 25


    [tex]\Rightarrow 0.83t^2 -10t+25=0[/tex]

    Use the quadratic equation formula and use the exact square root value.
  10. Jul 23, 2008 #9
    Duh! Thanks :)

    Now I get this though:


  11. Jul 23, 2008 #10


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    Yes, now just use your calculator to find which one gives you 3.5 seconds, and use that exact value in your calculations.
  12. Jul 24, 2008 #11
    I found that it is this one:


    where v(x) = 10 - 1.66t

    and the calculator still gives me the approximation of [tex]\sqrt{17}[/tex] which is 4.12310562562 and not [tex]\sqrt{17}[/tex] itself.

    I think this is because I have 1.66 in the function.. by the way, I see a "convert to decimal" function on my calculator.. but not a "convert to fraction" function. Can 1.66 be expressed in terms of a fraction/in "fractional form", if you will?

    I shouldn't be bothering with all this calculator stuff.. all I'd like to know is how I am supposed to arrive at [tex]\sqrt{17}[/tex] as the exact answer, as the book says. After all, with something like 4.12310562562 on my display, there is no way I can know that to be [tex]\sqrt{17}[/tex] off the top of my head.
  13. Jul 24, 2008 #12


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    at time



    see where the 1.66 cancels out? Then just simplfiy.
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