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Homework Help: Speed Average Velocity Change Question

  1. Sep 30, 2006 #1
    This is my first post here, I am going for high college physics, this is my second physics class in my life. (One in highschool and this is Physics 100 -- Survey of Physics.)

    We do all our homework online, so here's the first one..

    A person walks first at a constant speed of 3.40 m/s along a straight line from Point A to Point B and then back along the line from B to A at a constant speed of 5.60 m/s. What is

    (a) her average speed over the entire trip?

    For average I simply did 3.4+5.6 = 9 Then I did 9/2 (Two different speeds) and I got 4.5.. but for some reason it's showing up wrong on my webassign. Is there something small I'm missing? Any help would be appreciated.

    Second Question

    A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 358 m in 168 s. What is her average velocity (both magnitude and direction) for the entire fall?

    So for the free fall v = (a)(t) and because a = g in this instance I did v = (9.8)(15.0) = 147 m/s

    So for the free fall velocity = 147

    The second part is confusing, The parachute is open so a != 9.8 because of the wind drag on the chute, does a = 1/2gt^2? Cause if it does I think I can figure it out.

    Thanks for your help in advance.
  2. jcsd
  3. Sep 30, 2006 #2

    Doc Al

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    Your thinking here is incorrect. Despite the temptation, you don't find average speed by adding two speeds and dividing by two (except under special conditions like constant acceleration). An exaggerated example will make the point: Say you travel one mile at a speed of 1 mile per year, then return at a speed of 1,000,000 miles per year. Is your average speed around 500,000 miles per year? No! You spend much more time going slower so your average speed is much slower!

    Instead of trying to average the speeds directly, use the relationship: average speed = distance traveled divided by time.

    Again, don't get hung up calculating velocities at various points. Use: average speed = distance traveled divided by time.
  4. Sep 30, 2006 #3
    Ah, that's right because speed = velocity when the direction is the same. I forgot that point, alright let me work it out.
  5. Sep 30, 2006 #4
    Sweet, I got it.

    a = d/t

    s = 625m+358/(15+168)

    625+358 = 983
    15+168 = 183

    987/183 = 5.37 == Right Answer.

    For this one

    I don't know time traveled or distance traveled so how would I go about using savg = dt
  6. Sep 30, 2006 #5

    Also this question is bringing up problems also:

    A rocket rises vertically, from rest, with an acceleration of 2.8 m/s^2 until it runs out of fuel at an altitude of 1300 m. After this point, its acceleration is that of gravity, downward. (Assume that the positive direction is upward.)

    (a) What is the velocity of the rocket when it runs out of fuel?
    (b) How long does it take to reach this point?
    (c) What maximum altitude does the rocket reach?
    (d) How much time (total) does it take to reach maximum altitude?
    (e) With what velocity does it strike the earth?
    (f) How long (total) is it in the air?

    For A

    I tried 1300/2.8 -- I got 464, I assumed that meant it took 464s to get to 1300m but v = at and (2.8)(464) = 1300 which obviously cancels them out, then I thought v = deltaX/deltaT deltaX being 1300 and deltaT being 464 and I got 20.31 which is also wrong..

    I really want to do good in physics, I understand the formulas I just don't understand how to apply them to certain problems.
  7. Sep 30, 2006 #6


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    Regarding B - use the expression for displacement of the rocket to obtain the time it takes to reach 1300 meters. Further on, plug that time into the equation of the velocity of the rocket. You now have the answer to A, too. For the other parts, use the equation of displacement for a free fall - meaning the only acceleration is gravity.
  8. Sep 30, 2006 #7
    I think what I am most confused on is how to find time? Is the time to 1300m really 464s? Or did I do that wrong?
  9. Sep 30, 2006 #8


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    You got it wrong. The equation of displacement equals x(t) = 1/2*a*t^2 . Plug in the acceleration and height into the equation and solve for t.
  10. Sep 30, 2006 #9
    That doesn't make sense to me, where do I plug in "height" I see where to plug in acceleration, but t^2 is not height.. unless you meant x(d) as in change in distance..
  11. Sep 30, 2006 #10


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    x(t) means: the height x in the moment t (i.e. the change of distance with respect to time). So, if we write 1600 = 1/2*2.8*t^2, we get the moment t from the equation, which is the moment the rocket reaches the height of 1600 meters.
  12. Sep 30, 2006 #11
    Well I figured out a and b

    A is 85.32
    B is 30.47

    I got A by doing:

    [tex]v^2 = 2(a) \Delta X[/tex]
    2a = (2.8)(2) which equals 5.6
    (5.6)(1300) = 7280
    Square Root(7280) = 85.32

    I got B by doing:

    85.32 = 0 + (2.8)t
    divide by 2.8
    You get 30.47
    so t = 30.47

    so C

    (c) What maximum altitude does the rocket reach?

    When it runs out of fuel at 1300m the acceleration becomes -9.8m/s and at that point the velocity is 85.32. So I tried to do:
    [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

    0 = 85.32 + 2*-9.8(x)
    -85.32 = -19.6(x)
    which is 4.35
    and 4.32^2 is 18.94 which is incorrect.

    So what did I do wrong on C?
    Last edited: Sep 30, 2006
  13. Sep 30, 2006 #12


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    Just use [tex]v = v_{0} + at[/tex], i.e. 0 = 85.32 - 9.81 * t , and solve for t to get the time it takes to reach the maximum altitide. Then, plug that time t into [tex]x = x_{0} + v_{0}t + \frac{1}{2}at^2[/tex], i.e. x = 1300 + 85.32*t + 1/2*(-9.81)*t^2 in order to get the maximum height.
  14. Sep 30, 2006 #13
    Okay, so

    -85.32/9.81 = 8.697


    [tex]x = 1300 + 85.32*8.697 + 1/2*(-9.81)*8.697^2[/tex]

    [tex]8.697^2 = 75.642[/tex]

    [tex]x = 1300 + 85.32*8.697 + 1/2*(-9.81)*75.642[/tex]

    Then 85.32 * 8.697 = 742.028
    Then 1/2(-9.81) = -4.905
    Then -4.905 * 75.642 = -371.024
    So x = 1300 + 742.021 + 371.024
    So x = 1670.997 which is the correct answer, thank you.
    Last edited: Sep 30, 2006
  15. Oct 5, 2006 #14
    For E.

    How would I start E, I know the answers of A, B, C, D.

    A: 85.32m/s
    B: 30.47s
    C: 1670.997m
    D: 38.47s
    E: ?

    (e) With what velocity does it strike the earth?

    I tried [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]
    [tex]v^2=85.32+2(-9.8) \Delta x[/tex]
    v^2 = 65.72(1670.997)
    v^2 = 109817.922
    v = 331.38 m/s

    Which is incorrect, since I don't know the time it takes from 1670-0 I can't use [tex]v = v_0 + a t[/tex].

    Any help or advice would be appreciated.
  16. Oct 5, 2006 #15


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    At the rocket's highest point what is it's velocity (it's not 85.32 m/s)?
  17. Oct 5, 2006 #16
    Using v = vi + at

    I got v = 85.32 + (-9.8)*(8.7)
    v = .06

    Is that correct?
  18. Oct 5, 2006 #17


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    Not quite. What does the rocket do once it reaches it's highest point?
  19. Oct 6, 2006 #18
    It stops, the velocity is 0. I was sitting in class a realized it, haha. I feel stupid ;).

    Anyways, I figured it out it was -180.91 m/s. :)

    Thanks for your help I'll probably need help on some other stuff so if I can't figure it out I'll post ;)
  20. Oct 6, 2006 #19


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    One gets used to that feeling, unfortunately my eureka moments usually happen just as I'm falling asleep :grumpy:
  21. Oct 6, 2006 #20
    For some reason this problem is really getting past me, I've tried a couple different things and still can't figure it out. I can't do savg = d divided by time because I know neither.
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