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Need Help On Uniform Motion Question please!

  1. Oct 1, 2007 #1
    Help me with this physics question PLEASE!?

    ok it goes:
    "Bob measures the motion of a toy car. It starts 4m east of his reference point and moves 4m east, 6m west, and 9m east. It takes 20 s to complete all these moves."
    and then it asks:
    a. find the total distance travelled, which I think is 19 m (4m +6m +9m). Am i right?
    b. Find the final position of the car. Is it 11m?
    c. Find the total displacement of the car. What?!!!
    d. Find the car's average speed. Uh, is average speed the same thing as average velocity? In the case it was, I think the answer is the same as e.
    e. Find the car's average velocity. Vave=total distance/total time = 19m/20s = 0.95 m/s (if my answer is right, are the significant digits right?)
    Thanks so much to anyone who is willing to help out a very confused physics pupil!
     
  2. jcsd
  3. Oct 1, 2007 #2

    Mentz114

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    Hi Cherryrocket.

    a) seems correct.

    b) the final position is 2 numbers, like x east and y north. Draw a checker-board diagram.

    c) I assume they mean the distance between the intial point and final point ( use Pythagoras)

    d) speed is a scalar, velocity is a vector. Your answer to e) looks OK for this one.

    e) Hmm. Work out the average velocity northwards, and the average velocity eastwards then combine to give a final vector average.
     
  4. Oct 1, 2007 #3
    thanks Mentz114 for responding!
    I have a question regarding part e)
    How can I separate north and east velocity when I only have one time (20s). I can understand if there was time intervals, but there isn't. So this is what I came up with:
    East Velocity = 4m+9m = 13m/20s = 0.65m/s
    West Velocity = 6m/20s = 0.33m/s
    And then I combined their averages: 0.65m/s + 0.33m/s = 0.98/2 m/s = 0.49 m/s is my final answer... but I don't know if it's right...
    And for part b), what do you mean by checker-board diagram? What I did was make a line graph, like the x-axis of a graph, and then started at 4 from my reference point of 0, jumped 4 to the right, jumped 6 to the left, and 9 to the right. And I ended at 11 as my answer. I am lost on this one...
     
    Last edited: Oct 1, 2007
  5. Oct 1, 2007 #4

    Mentz114

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    Whoops, I misread your question, you never move north/south so you don't need the 2D graph.

    b) you just need to say how far east or west the car is.

    You've nearly got e), the thing is that east is in the opposite direction to west, so you should subtract your 2 averages. The answer is 0.32 east (please check!).

    Apologies for the confusion ( and dyslexia).
     
  6. Oct 1, 2007 #5
    Thanks a bunch, no need to apologize you are helping me nonetheless! So then, the answer for b) would be 11m EAST because I ended up in the positive side of the x-axis I drew, right?
     
  7. Oct 1, 2007 #6

    Mentz114

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    That's right. It's all the information you need to locate the car is there.
     
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