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Help on a Relative Motion Problem

  1. Mar 16, 2005 #1
    Help on a Problem!!

    Hi there everyone. I am completely lost on how to go about answering this question :cry:

    A sunbather, drifting down the Grand River on an inner-tube, dives off the tube as it passes under the high bridge of Elora Gorge. She swims against the current for 15 mins before turning and swimming downstream with the same effort (ie: same speed relative to the water). When she finally reaches her tube, she is 1.0 km downstream from the high bridge. What is the current of the water?

    If anyone can help solve this question it would help A LOT!!
     
    Last edited: Mar 16, 2005
  2. jcsd
  3. Mar 16, 2005 #2
    :cry: :cry: :cry: :cry: :cry:
    can nobody help me??
     
  4. Mar 16, 2005 #3
    I didn't follow this all the way through, but I believe you should be able to solve it with algebra and careful book keeping:

    Facts:
    D(Inner Tube) = 1 km, and therefore 1 = (rate of river) * total time
    (D1) = distance traveled by girl in 15 minutes
    (D2) = distance traveled by girl in remaining time
    D1+D2 = D(Inner Tube)
    D1 = [(rate of girl) - (rate of river)] * 15 minutes
    D2 = [(rate of girl) + (rate of river)] * (total time - 15 minutes)
     
  5. Mar 16, 2005 #4

    xanthym

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    From the problem statement:
    {Positive Direction} = {Downstream}
    {River Current Speed Relative To Shore} = R
    {Swimmer Speed Magnitude Relative to River} = S
    {Swimmer Upstream Speed Relative to Shore} = R - S
    {Swimmer Downstream Speed Relative to Shore} = R + S
    {Time Spent Swimming Upstream} = tu = (15 min) = (0.25 hr)
    {Time Spent Swimming Downstream} = td

    From the swimmer's data:
    {Swimmer Upstr Speed}*{Time Upstr} + {Swimmer Downstr Speed}*{Time Downstr} = (1 km)
    ::: ⇒ (R - S)*tu + (R + S)*td = (1 km)
    ::: ⇒ (R - S)*(0.25 hr) + (R + S)*td = (1 km)
    ::: ⇒ (0.25)*R - (0.25)*S + (R + S)*td = (1) ::: Eq #1

    From the tube's data:
    R*(tu + td) = (1 km)
    ::: ⇒ R*((0.25 hr) + td) = (1 km)
    ::: ⇒ td = (1/R) - 0.25 ::: Eq #2

    Placing Eq #2 into Eq #1:
    (0.25)*R - (0.25)*S + (R + S)*{(1/R) - 0.25} = (1)
    ::: ⇒ (0.25)*R - (0.25)*S + 1 - (0.25)*R + S/R - (0.25)*S = 1
    ::: ⇒ 1 + S/R - (0.5)*S = 1
    ::: ⇒ S/R - (0.5)*S = 0
    ::: ⇒ S - (0.5)*S*R = 0
    ::: ⇒ 1 - (0.5)*R = 0
    ::: ⇒ (0.5)*R = 1
    ::: ⇒ R = (2 km/hr)


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    Last edited: Mar 16, 2005
  6. Mar 17, 2005 #5
    Thanx!! :smile:
     
  7. Mar 17, 2005 #6
    Thank you so much! I'm still slightly confused though. I get how you got the data from the problem statement...and I understand the substitution of the 2nd Eq'n into the 1st.

    Can you explain to me how you got the 2 initial equations:

    1) {Swimmer Upstr Speed}*{Time Upstr} + {Swimmer Downstr Speed}*{Time Downstr} = (1 km)

    2) R*(tu + td) = (1 km)

    Like is there some kinematic eq'n that can help me understand how you got them??
     
  8. Mar 17, 2005 #7

    xanthym

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    Science Advisor

    Both Equations #1 & #2 above for which you wanted clarification are applications of the basic kinematic relationship:
    {Distance} = {Speed}*{Time} ::: Eq #3 (Valid for constant speed over given time interval)
    In each case above, an entity's speed is multiplied by the total time during which it maintained that (constant) speed to determine the distance traveled. Equation #1 is actually composed of two distance segments; however, each distance segment is calculated from this basic kinematic equation (Eq #3).


    ~~
     
    Last edited: Mar 17, 2005
  9. Mar 19, 2005 #8
    Thanks again!!!
     
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