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Kinematics (Vectors, fairly easy stuff)

  1. Jul 19, 2006 #1
    Can someone please help me out with some of these questions?

    1) A scientist runs to the east at 1.2m/s when he suddenly realises he forgot his chemicals at home. He then dashes at 4.6 m/s [North]. What is the scientists change in velocity?

    So far I have 4.6 m/s - 1.2 m/s = 3.4, does anyone know what I am supposed to do with the directions?

    2) (sorry I don't know how to put in the degree sign infront of the numbers, I will use ' instead) The velocity of a plane relative to the air is 320 km/h [E 35' S]A wind velocity relative to the ground is 75 km/H [E]. What is the velocity of the plane relative to the ground.

    Again, in this one it's the direction (or angle) that makes me unsure of my answer, I am using the Vog = Vom + Vmg (velocity of object realtive to ground = velocity of medium relative to ground + velocity of object relative to medium) but the fact that the plane is flying at an angle of 35 degrees confuses me.

    help would be appreciated :biggrin:
     
  2. jcsd
  3. Jul 19, 2006 #2

    berkeman

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    Staff: Mentor

    -1- To include the directional part, you need to draw the velocity vectors, and do a graphical subtraction. Draw an arrow 1.2 units long pointing to the right (east), and at the tip of that arrow, draw another arrow 4.6 units long pointing up (north). The addition of the two vectors is the hypotenuse of the triangle (start at the beginning of the first vector and end at the pointy end of the second vector). Do you have an idea of how you would construct the difference (subtraction) between these two vectors?

    -2- In this one you add the plane's velocity vector and the wind vector. Use the method from above.
     
  4. Jul 19, 2006 #3
    I think for the angle I would have to use tan-1(hyp)? (-1 being in the degree mode) right?
     
  5. Jul 19, 2006 #4

    berkeman

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    Actually, an easier way to do -1- would be to express each vector in rectangular coordinates, and do a simple subtraction. Have you seen this notation before? Assume that the horizontal left-right axis on your paper is the x-axis, and the vertical up-down axis is the y axis. Then each velocity vector can be expressed as the sum of its x and y components:

    V = (Vx,Vy)

    So for example, the first velocity vector would be V1 = (1.2m/s,0).

    What would V2 be in this notation? What do you get for V2-V1 if you subtract each component of the vector?
     
  6. Jul 19, 2006 #5
    V1 = (1.2m/s, 0)
    V2 = (0, 4.6)

    Well...if I put those in the pythagorean equation, I get a final answer 4.9 and then I put it in tan-1 and I get approximatly 79 degrees...am I on the right track or am I still off?

    Edit: oh wait, I see, am I supposed to subtract the x-component of V1 with the x-component of V2 and then do the same for the y-component and that would be my delta V?
     
  7. Jul 19, 2006 #6

    berkeman

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    Correct for V1 and V2. Now just subtract them to get V2-V1. You'll see that the difference vector basically says that the x-motion stops and the y-motion starts. You can also see that if you add the two vectors instead, you get that hypotenuse vector I mentioned. I think you can also see that subtracting the V1 vector is like turning it around (flipping it horizontally in this case) and adding that to V2. The difference vector in this case is the hypotenuse of the triangle you get by flipping V1 horizontally, moving it to the right so that its tip touches the bottom of V2, and then taking that hypotenuse. Make sense?
     
  8. Jul 19, 2006 #7
    Yes, thanks, it makes sense now. For number two would I do the same? Since I have an angle this time, I think I need to do something like

    V1 = (320km/h SIN35, 320km/h COS35)
    V2 = (75, 0)
     
  9. Jul 19, 2006 #8

    berkeman

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    That looks right. What you are doing is finding the x and y (rectangular) components of the vector V1. It's called converting from polar coordinates to rectangular coordinates. In polar coordinates, you are given magnitude and direction (as an angle from one of the axes). In rectangular coordinates, you are given the x and y components of the vector, and the magnitude is the hypotenuse of those two components, and the angle is the [tex]tan^{-1}[/tex] of the y-component divided by the x-component.

    There should be a section in your textbook that talks about converting between coordinate systems. As you can see, using rectangular coordinates makes adding and subtracting vectors easy. Good work!
     
    Last edited: Jul 20, 2006
  10. Jul 19, 2006 #9

    berkeman

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    One other thing -- I didn't check the signs of the components of V1 (I don't understand the notation of East-South, etc., and I don't have a diagram to help me figure it out). But remember that if the x-component points left, it will have a negative magnitude. And if the y-component points down, it will have a negative magnitude.
     
  11. Jul 19, 2006 #10
    Thank you so much for all that you've done, you've really been a big help.
     
  12. Jul 19, 2006 #11

    berkeman

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    You're welcome, glad to help. Say, can I ask for a return favor -- I posted a question about your reply in the 1337 speak thread. Can you check it out? It's at the bottom of the thread:

    https://www.physicsforums.com/showthread.php?p=1041249#post1041249

    Thanks, -Mike-
     
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