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Kinematics and Motion

  1. Jan 23, 2014 #1
    1. The problem statement, all variables and given/known data

    I currently have a few problems I am working on for my calculus based physics class. I'm duel enrolling in both calc based physics and calc at this moment.

    |||The first problem is The figure (figure 1) shows the motion diagram, made at two frames of film per second, of a ball rolling along a track. The track has a 3.0-m-long sticky section.

    Here is figure 1
    200983023371863387272238927625093.jpg

    The questions pertaining to this particular problem are as follows.
    A) Make a position-versus-time graph for the ball.

    B) What is the change in the ball's position from t = 0 s to t = 1.0 s?

    C) What is the change in the ball's position from t = 2.0 s to t = 4.0 s?


    ||| The Second problem A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 32m/s2 for 35s , then runs out of fuel. Ignore any air resistance effects.

    A) What is the rocket's maximum altitude?

    The answer I got that was correct was 84 km.

    B) How long is the rocket in the air? Express in seconds with two significant figures.


    ||| Problem 3

    A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. You can ignore any effects of air resistance.

    A) What was the rocket's acceleration during the first 16 s?

    B) What is the rocket's speed as it passes through a cloud 5100 m above the ground?



    2. Relevant equations

    Problem 1 Equations?

    Problem 2 Equations?

    Problem 3 Equations?




    3. The attempt at a solution

    Problem 1: I am a little disoriented by this problem and I am not sure where to start.


    Problem 2: I have attempted this problem and this is all I've been able to do. What's the next step?
    32m/s2*35 s = 1120 m/s
    .5*32m/s2*(35)s2+.5*(32*35)s2/9.81 = 83534.7m/1000 = 84 km

    I calculated the time after it stops accelerating that it is within the air to be 114.4 seconds; do I now add the 35 seconds to this to get my answer? Mastering Physics is saying 149 is not correct.

    Where do I go from here?



    Problem 3: I attempted but my work got me 25.5 m/s2 which it said was wrong. I'll edit the equations and work I attempted tomorrow. It's gotten late and I need a fresh brain to properly input these figures.


    I can do these problems I know I can... I just am not sure which formula to use for which part of each question. Explaining that would be great. Mostly just giving me the appropriate kinematic equations would be a enough; Along with a break down of what each symbol or variable stands for... Please anyone help me with the appropriate equations to use for these problems!
     
  2. jcsd
  3. Jan 23, 2014 #2

    CWatters

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    Hint: So what is the time interval between each image of the ball?

    Hint: Try writing the time next to each image of the ball.
     
  4. Jan 23, 2014 #3
    According to the problem there are 2 frames shot per second so the time interval between each image would be 0.5 seconds; I understand that but I don't know where to go from here. I tried eyeballing to get the answer but masteringphysics is being stubborn or I have to do it mathematically.


    For the sticky section I've determined all the other answers and plotted a graph for position versus time. However, I can't seem to calculate Acceleration...

    Determine the ball's acceleration on the sticky section of the track.
     
    Last edited: Jan 23, 2014
  5. Jan 23, 2014 #4

    BvU

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    Speed = change in position / change in time

    from 0 to 1 s the speed v1 is ? It is constant because ..?
    from 2 to 3 s the speed v2 is ? It is constant because ..?

    On the sticky section the speed changes from v1 to v2. The change is v2 - v1.

    Acceleration = change in speed / change in time
     
  6. Jan 23, 2014 #5

    BvU

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    The first formula you don't mention under relevant formulas I recognize as v = at.
    It is over three times the speed of sound, but never mind.


    The second formula you don't mention under relevant formulas looks a bit sloppy. Either try to treat numbers and units separately or keep them together more meticulously. It will save you errors.

    You do an energy balance and work out g h0 + 1/2 v0^2 = g hmax + 1/2 vmax^2 with h0 = 1/2 a t^2, v0 = a t and vmax = 0.

    Ok, gets you 84000 m and NOT 84000m/1000 !

    (Both formulas deal with part A. I misread the OP and reacted too hastily)
     
    Last edited: Jan 23, 2014
  7. Jan 23, 2014 #6

    BvU

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    Now for part B. After 35 seconds, the height can be expressed as a function of time: h0 + v0 (t-35) - 1/2 g (t-35)^2 (same h0, v0 as in A, t = 0 at firing when the rocket starts being in the air). Putting height(t) = 0 yields a quadratic equation in t
     
  8. Jan 23, 2014 #7

    BvU

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    What do you have already ? It looks as if there are no relevant formulas at all to be found, which I don't believe.
    Calculus based physics should include some treatment of the subject of uniform motion, uniformly accelerated motion and projectile trajectories ?
    If not, most physics books do exactly what you ask: they try to provide you with a set of equations to apply, together with an explanation of what the symbols or variables stand for. You can't expect to get a much better general treatment from PF (but if you have specific, pointed questions, PF is a good bet!) ;-)
     
  9. Jan 24, 2014 #8
    Okay, so for the first problem I've finally managed to figure it out using references in my book and seeing my instructor for help.
    Problem 1
    B) The change in the ball's position from 0 to 1 second is 4.0 m

    C) The change in the ball's position from t = 2.0 seconds to t = 4.0 seconds is also 4.0 meters.

    Realizing that each picture of a particle position was shot every 0.5 seconds helped me on this and then it was just processing it visually to get the answers. Sometimes after doing physics for 6 hours my brain has a hard time looking at a simple graph!

    A= vf-vi/time thank you for the help!!

    Problem 2

    For problem 2 I used something like this... for part A but as for part B I'm at a loss to calculate how many seconds it remains in the air...

    h1=.5*32*35^2

    v1=32*35

    to apogee
    t2=32*35/g

    h2=h1+.5*(32*35)^2/g

    h2=.5*32*35^2+.5*(32*35)^2/9.81

    Giving me roughly 84000 meters which I then converted to KM


    As far as part B the answer is 280 seconds; However, I am not sure how to achieve that ...



    Problem 3

    As far as this problem goes I asked my instructor and they said they had to omit it because they haven't yet taught us how to do problems such as these. They said it had something to do with two variables two unknowns and they took 30 minutes trying to explain how to do it only to realize it was very convulsed and overly difficult for where we are in the class.
     
  10. Jan 24, 2014 #9

    CWatters

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    Break the problem down into 3 sections. 2 on the way up and 1 on the way down. Apply the SUVAT equations of motion to each section to solve for the time taken for each section. Add them up. I haven't checked if 280 is the right answer.
     
  11. Jan 24, 2014 #10

    BvU

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    You can follow CWatters, or you can follow post #6.
    Do you understand what I mean to say in post #5 ?
    I sincerely hope that what you type is a short version of what you've written down, something like
    being shorthand for h1 = 1/2 * a * t^2 = 1/2 * (32 m/s^2) * (35 s)^2 =.5*32*35^2 m (as in post #5, only I called it h0, the starting height for the next part of the exercise).

    Now, just to encourage you: following post #5 does give you the right answer.

    But don't be fooled: the object of the exercise is NOT that you get the right answer. The object is that you understand the subject matter so good that you can find and follow the correct way to get the answer.

    Keep at it and be welcome to start a new thread when you re-attack problem 3 after you've been handed enough know-how to be ablee to deal with it.
     
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