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Homework Help: Apparently simple canoe problem - or not?

  1. Feb 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A canoeist heads north for 1.2km on a lake and then heads west for 1.8km. Her speed is steady at 5m/s.
    a) How long does it take her?
    b) What is her average velocity?

    2. Relevant equations

    3. The attempt at a solution

    Total of 600s but the answers suggested 0.60h

    The average velocity was suggested as 3.6km/h or 1m/s but throughout her trip she held a constant speed of 5m/s.

    So there is concern for the suggested answer for both a) and b)
  2. jcsd
  3. Feb 14, 2007 #2
    Pivoxa, what book are these coming from? Are we outsourcing textbook writing these days as well? The first answer I agree with you.
    If you assume they meant 5km/hr, .6hr fits. See if the second answer makes sense using this figure and calculating net displacement.
  4. Feb 16, 2007 #3
    They are looking for the average speed in b) so if we assume an average velocity of 5km/hr (which what they used in a)) then b) should be 5km/hr not 3.6km/hr.

    I am using an Austrailan textbook from Nelson publishing company. There are many, many problems in it and errors may crop up at times.
  5. Feb 16, 2007 #4
    well, needs a good dose of editing.............

    Remember the definition of speed and velocity are very different in physics than in everyday language. For instance if you drive around a racetrack at an ave speed of 250Km/hr, your velocity between the green and checkered flags is still zero, 'cuz you haven't gone anywhwere, ie net displacement is zero. So in problem above, same applies, calculate actual displacement from starting point then divide by time to get velocity, which is a vector quantity (has direction associated with it). Speed is a scalar quantity, no direction implied.
  6. Feb 16, 2007 #5


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    First thing to do is draw a picture and think about the info given in terms of vectors and scalars...

    The resultant of the two directions is 2163m - this is what you have to equate your average speed to - ie. 2163/5=0.6h
    Last edited: Feb 16, 2007
  7. Feb 16, 2007 #6


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    No one can canoe at 5 m/s! I feel sure 5 km/h was intended. Then the total time was 0.60 hours. Remember that velocity, unlike speed, is a vector quantity. The magnitude of the velocity will be the straight line distance from starting point to ending point (use the Pythagorean theorem). You can use arctangent to get the angle.
  8. Feb 16, 2007 #7

    D H

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    A very few select people can, for short distances. World-class times for the C-1 200 meter sprint are below 40 seconds (5 m/s), C-1 500 meter sprint, about 1:46 (4.7 m/s), K-1 500 meter sprint, about 1:38 (5.1 m/s).

    No one can paddle at 5 m/s for any sustained period of time. Certainly not over a distance of 3 km.
  9. Feb 16, 2007 #8
    I see. velocity is displacement/time
  10. Feb 16, 2007 #9
    And understanding this is a key to understanding much of physics--vectors have direction. scalars don't. Forces are vectors. Consider a merry go round, you can apply force in many directions and not get it to budge. Thats why these distinctions matter.
  11. Feb 16, 2007 #10
    What are you trying to illustrate with your example? Could you be clearer? That forces may balance and cancel hence get 0 displacement and 0 velocity? Speed would be 0 as well though.
  12. Feb 17, 2007 #11
    Sorry, just trying to illustrate the importance of direction and why forces must have directions attached. You could push up, push down, push inwards toward the center, and the merry-go-round might deform, but not move. Apply a tangentially directed force to its axis of rotation and Voila!
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