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Speed as a function of time

  1. Jul 17, 2015 #1
    Hi everyone. I am trying to understand what is the speed of my dog with which he is jumping out of the bath tub. I calculated to -4.9 m/s which is half of g.

    Code (Text):

      ^   |
      h   |
      v   _____B___
    Here are some details that are known:
    h = 53cm
    Vi = 0
    g = -9.8 m/s^2
    d = 140cm

    I also appreciated that it took him to reach point B from point A in roughly 1 second (that is, from the moment he jumped from A, middle of bath tub and reached ground floor at point B, 1 second passed.

    I've tried v = d/delta t, where d = -4.9 so -4.9/1 = -4.9 m/s

    Is this correct?

    P.S.: i do not know the angle
  2. jcsd
  3. Jul 17, 2015 #2


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    By the Pythagorean theorem, then, Your dog moved a distance [itex]\sqrt{53^2+ 140^2}= 150[/itex] cm in 1second (I cannot see how you got a distance of 4.9 meters!) so 150 cm/s= 1.5 m/s.

    It doesn't makes sense to say "4.9 m/s" is "half of g" because g is (approximately) 9.8 m/s^2, an acceleration not a speed.
  4. Jul 17, 2015 #3
    There must be an angle.Cause without any angle dog cant jump directly forward.
    Lets assume that angle 60 degree.Now we need to consider.Inital velocity which lets call it V.V has two components here ##V_x## and ##V_y##.Now ##V_x## component will determine the distance from inital point.##V_x=d/t## , ,d=Vtflight.Now here we know d and t so we can calculate ##V_x##.Theres 60 degree so we can calculate other components easily
    (Use meter to calculate distance)
  5. Jul 18, 2015 #4
    Well I do not really care about the angle since all I'm interested in is the magnitude.
    For -4.9 m/s (for those of you who are very precise: roughly -4.9 m/s) and I think I now realize that I got rid of the units assuming that I can do it whenever I got a multiplication with 0 (see Vi):
    d = \vec{V_{i}} \frac{m}{s} * \Delta t \ s + \frac{1}{2} * \vec{a} \ \frac{m}{s^{2}} * \Delta t ^{2} \ s \\ \Rightarrow
    d = 0 + \frac{\vec{a}}{2} \ \frac{m}{s^{2}} * \Delta t ^{2} \ s \\ \Rightarrow
    d = \frac{\vec{a}}{2} \ \frac{m}{s} * \Delta t ^{2} \Rightarrow d = \frac{\vec{a}}{2} \ \frac{m}{s} * 1 \\ \Rightarrow d = -4.9 \frac{m}{s}[/tex]
  6. Jul 18, 2015 #5
    Theres must be an angle.
    And your physics is not correct.
    You need to think its jumbed.
    First think d is a distance properity.Not speed.
    You said d=4.9 m/s which its not correct.Your equations is not true.


    I am trying to explain the idea of physics in the picture.Which your dog follow the same trajectory I guess.
  7. Jul 18, 2015 #6
    Yes, I agree there is an angle. What I said is that I am not interested in finding it because I only want the magnitude.

    Yes, I just realized it's really screwed up and I cannot derive like in maths but have to stick to making sense out of it. Of course distance is measured in meters not unit/time.

    I will come back later with a full solution to it. BTW what software did you use to draw the graph?
  8. Jul 18, 2015 #7


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    To find the Y velocity, you only need to calculate from the bottom of the tub to the highest point the dog reached (assuming that the tub floor is the same height as the ground). If we just use the time of the jump, we can use an easier equation: V = Vi + At. Final velocity is zero since we are only calculating to the point where the dog's Y velocity is zero, which is halfway through the jump. Time is 0.5 seconds.
    0 = Vi + (-9.81)(0.5)
    Vi = 4.9 m/s

    To find the X-velocity, we can also use a simpler equation than what you used if we assume air resistance is negligible: X = Xi + Vit
    1.4 = 0 + Vi(1)
    1.4 = Vi
    Vi = 1.4 m/s

    By the Pythagorean theorem: V = √1.42 + 4.92
    V = 5.09 m/s
    Last edited: Jul 18, 2015
  9. Jul 18, 2015 #8


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    Standby, I'm editing my above post.
    Edit: Editing done.
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