Magnitude of average velocity and average speed

  • Thread starter J.live
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Homework Statement


An electron moves in the positive x-direction a distance of 2.44 m in 3.10 *10-8 s, bounces off a moving proton, and then moves in the opposite direction a distance of 1.68 m in 3.53*10-8 s.



The Attempt at a Solution



to find the average velocity I used the distance formula to find magnitude divide it by the c

Sqrt (2.44)^2+(1.68)^2 / (3.53*10^8 - 3.10*10^8) ?


for the average speed

(2.44 - 1.68) / (3.53*10^8 - 3.10*10^8)?


Can someone explain it by showing some work? I keep getting the wrong answer.
 

Answers and Replies

  • #2
cepheid
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for the average speed

(2.44 - 1.68) / (3.53*10^8 - 3.10*10^8)?


Can someone explain it by showing some work? I keep getting the wrong answer.

This part is wrong for sure. The average speed would be the total distance travelled (without regard for direction) divided by the total time elapsed. The total time elapsed would be the SUM of the two time intervals, not the difference between them, right?
 
  • #3
cepheid
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As for the first part of the question, your method for finding the net displacement is also wrong. That Pythagorean distance formula you used does not apply, because the two vectors aren't perpendicular. They are anti-parallel (meaning one is in the +x direction, and one is in the -x direction). So they lie along the same line, making this effectively a 1D problem. To add up the two vectors, all you end up having to do is subtract the two distances. Draw a picture and you'll see what I mean.

A more intuitive way to arrive at that answer is to think about what the net displacement actually means physically: it is how far from its starting point the particle ends up.

My comment about the time interval also applies here as well.
 
  • #4
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Thank you.
 

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