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Average speed question: forwards then backwards

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    An object starts at the origin, position A; then travels 20 m to position B; then returns to position A. From A to B the average speed is 10 m/s; on its way back the average speed is 6 m/s. What is the average speed for the entire trip?

    2. Relevant equations
    average speed = distance/time
    distance is the "length" covered. This is not - according to my understanding - the same as velocity, which is displacement/time. In this example, the velocity would by 0 m/s, because the numerator, the displacement, is zero, as the object has returned to the origin.

    3. The attempt at a solution
    From A to B the distance is 20 m and the average speed is 10 m/s. Therefore:
    20 m/time = 10 m/s and time = 20 m/ 10 m/s = 2 s.
    From B to A the distance is 20 m and the average speed is 6 m/s. Therefore:
    20 m/time = 6 m/s and time = 20 m/ 6 m/s = 3.33 s.
    Overall, the journey from A to B and back again takes (2 + 3.33) s = 5.33 s.
    The total distance covered (not displacement) is 40 m (20 m from A to B and 20 m from B to A).
    Therefore, the average speed = distance/time = 40 m/ 5.33 s = 7.50 m/s (approx.).

    I have changed the numbers from the textbook to see if I was making some type of numerical error, but I have applied the same reasoning. The textbook has a different value from what I obtain (it is lower). Where am I going wrong?

    Any help appreciated, many thanks.
     
  2. jcsd
  3. Nov 9, 2014 #2

    HallsofIvy

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  4. Nov 10, 2014 #3
    In the textbook the value from A to B is 8.8 m/s and from B to A it is 6.6 m/s. They give the answer as something like 2.6 m/s. I got 7.54 (to 2 d.p.):
    40 m / [(20/8.8) + (20/6.6)] s = 7.542857143 m/s.
     
  5. Nov 10, 2014 #4
    2.6 m/s does not make sense. It cannot be less than the 6.6 m/s, the lowest of the two average speeds.
     
  6. Nov 16, 2014 #5
    Thanks for the reply. I guess it's just an error in the textbook.
     
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