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Homework Help: Relative Velocity

  1. Aug 23, 2010 #1
    A swimmer is training in a river. The current flows at 1.99 metres per second and the swimmer swims upstream a distance of 153 metres before swimming back to the starting point. If the total time for the swim is 192.0 seconds, what is the swimmer's speed relative to the water?

    I've been stuck on this question for a while now just can't seem to get it out? Tried rearranging equations but still can't get to the answer of 2.94 m/s?

    Any help would be greatly appreciated thankyou
  2. jcsd
  3. Aug 23, 2010 #2


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    Homework Helper

    Show your work, please.

  4. Aug 23, 2010 #3

    Current = 1.99 m/s
    Total time = 192 s
    Total distance travelled = (2x153) = 306 m

    I've used the total time taken taken (192s) to be equal to the formula for the time taken upstream and downstream which looked like this.

    192 = (153/ (x + (-1.99)) + (153/(x + 1.99)

    rearranging and multiply out i've got down to

    192 = 306x / x^2 - 3.9601

    not sure how to solve for x any help??
  5. Aug 23, 2010 #4
    So far so good.

    That's not right. Try again, you'll end up with a standard quadratic equation. Tip: multiply each fraction by the denominator of the other fraction to obtain a common denominator.
  6. Aug 23, 2010 #5
    That is what i had done cross multiplied to get that equation?

    Is the denominator then added to get 2x^2-7.9202 i'm oblivious of where to go from hear very lost atm?
  7. Aug 23, 2010 #6
    i really need some clarification on this question it is doing my head in why i can't work it out, there is something that i am not taking into consideration but i don't know what it is please some help is required!!!!!!
  8. Aug 23, 2010 #7


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    Homework Helper

    Did you mean


    ? Then do not forget the parantheses: 192 = 306x / (x^2 - 3.9601).
    Multiply the equation with the denominator, do the multiplication, arrange the equation so all terms are at one side, solve the quadratic equation.

  9. Aug 24, 2010 #8
    thanks for the help i resolved the question today what a relief
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