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Vectors: Displacement, average velocity & speed

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  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data
    A plane flies 413 km east from city A to city B in 49.0 min and then 814 km south from city B to city C in 1.70 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).

    2. Relevant equations
    so velocity is = to change in displacement over change in time

    3. The attempt at a solution
    For a, the magnitude of displacement is (413km^2+814^2)^.5 pythagorean theorem. is 912.8 km right?
    For b, the direction of the plane's displacement is tan^-1 (814.2/413). but this angle 63.104 must be expressed of magnitude less than 180 degrees measured from +x direction (east). So how would you do that? Because 63.104 degrees is positive and not right, do you just minus 180 degrees? Then is it just -116.902 degrees? Yeah I guess so.
    For c, the magnitude of average velocity is displacement 912.8km/ 2.52 hours to get 362.2222 km/hour.
    For d, is the direction of avg velocity just the same as the answer to b, is it just -116.902 degrees also?
    For e, isn't avg. speed just 486.9 km/hour?
    Basically I have got answers but am really unsure if I am doing it right or wrong. For all I know, I am completely missing the point. Thanks again.
     
  2. jcsd
  3. Jan 21, 2017 #2
    Is the answer to d and b the same?
     
  4. Jan 21, 2017 #3
    I think it is common for angles to be considered positive as you rotate CCW from the +x axis. So I would think that your angle would be -63.1 degrees. But it all depends on who defined it.

    I think your solution looks correct.
     
  5. Jan 21, 2017 #4

    gneill

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    Staff: Mentor

    Subtracting 180 degrees would reverse the direction of the vector. You can only add or subtract multiples of 360° if you want to retain the direction.

    In this case you've lost the sign of the north-south displacement. When the plane traveled southwards from B, what sign should you assign the displacement?
    Your answer to (a) is good. Fix up the answer to (b) by setting the correct sign on the displacements. You've got the right ideas for the rest.
     
  6. Jan 21, 2017 #5
    Wait, is (a) correct, then why did you say that I should change the sign of the displacement?
     
  7. Jan 21, 2017 #6
    oh, ok, so the direction is actually -63.098 degrees because tan ^-1 (-814.2/413).
     
  8. Jan 21, 2017 #7

    gneill

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    Staff: Mentor

    The signs of the components "disappear" when you square them using Pythagoras. So the sign didn't affect the result.
     
  9. Jan 21, 2017 #8

    gneill

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    Staff: Mentor

    FYI, I changed the title of this thread to make it descriptive of the actual problem. By the forum rules, too vague or too general titles can be grounds for thread deletion, so please take the time to formulate thread titles that are descriptive of the question being asked.
     
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