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Speed, Velocity and Average Acceleration

  1. Sep 6, 2009 #1
    1. At t = 0, an automobile traveling north begins to make a turn. It follows one-quarter of the arc of a circle of radius 9.3 m until, at t = 1.60 s, it is traveling east. The car does not alter its speed during the turn.

    (a) Find the car's speed.

    (b) Find the change in its velocity during the turn.

    (c) Find its average acceleration during the turn.

    2. speed = distance traveled/time traveled
    change in velocity = vf - vi
    average acceleration = change in velocity/change in time


    3. I am a little confused at how to begin solving this problem. I know that the first vector going north is going to be the initial velocity = vi and the second vector is going to be the final velocity = vf. the initial time = 0 sec and the finald time is = 1.60 s. Distance traveled is = 9.3 m. I am so lost at where to start. Can someone please guide me in the right direction to solving this problem?
     
  2. jcsd
  3. Sep 6, 2009 #2

    kuruman

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    Since the car is traveling at constant speed, it covers equal distances in equal times, so distance traveled is speed multiplied by time. You know that the time for a quarter-turn is 1.60 s. What is the distance traveled? It is not 9.3 m, it is one-quarter of a circle of radius 9.3 m. This should get you started.
     
  4. Sep 6, 2009 #3
    So would the distance traveled in 1.60 s be 9.3 m -0.25 = 9.05 m
    would speed be = 9.05 m/1.60 s = 5.65 m/s
     
  5. Sep 6, 2009 #4

    kuruman

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    What is the circumference of the circle? One-quarter of that is what you are looking for.
     
    Last edited: Sep 6, 2009
  6. Sep 6, 2009 #5
    I have no idea what the circumference of the circle is..It doesn't state that in the problem. It only gives the radius of the circle of 9.3 m
     
  7. Sep 6, 2009 #6

    kuruman

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    It is something you are supposed to know. Please memorize this formula. It is extremely handy when doing physics. The circumference of a circle is

    C = 2 π r

    where r is the radius.
     
  8. Sep 6, 2009 #7
    C = 2 pi r = 2 * 3.14 * 9.3 m = 58.4 m

    58.4 m - 0.25 = 58.15 m

    to get speed would it be = 58.15 m/ 1.60 s = 36.3 m/s
     
  9. Sep 6, 2009 #8

    kuruman

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    If you are jogging on a circular track such that one lap is 58.4 m, what is one-quarter of the way around the track? It is certainly not 58.15 m because that is almost a complete lap.
     
  10. Sep 8, 2009 #9
    Would it be 58.4 m * 0.25 = 14.6 m
    speed = 14.6 m/1.60 s = 9.12 m/s
     
  11. Sep 8, 2009 #10

    kuruman

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    That would be correct. Now for part (b), you need to write the initial and final velocities as vectors in unit vector notation. The change in velocity is the difference taken as Final velocity minus Initial velocity. I assume you know how to subtract vectors.
     
  12. Sep 8, 2009 #11
    So would the change in velocity be = 9.12 m/s - 0 m/s = 9.12 m/s because the initial velocity is 0 m/s?
     
  13. Sep 8, 2009 #12

    kuruman

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    The car is initially moving north. Its velocity is not zero. How would you write the initial velocity as a vector?
     
  14. Sep 8, 2009 #13
    So to find the initial velocity would you take 9.3 m^2 + 9.3 m^2 = 172.98 m and take the sqr rt = 13.2 m

    to get velocity = 13.2 m/1.60 s = 8.22 m/s <---initial velocity?
     
  15. Sep 8, 2009 #14

    kuruman

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    This problem is in two dimensions. Any 2-D vector A can be written in unit vector notation as

    [tex]\vec{A}=(something) \hat{i} + (something \ else)\hat{j}[/tex]

    What must you put in the above expression for "something" (or x-component) and "something else" (or y-component) to get the initial velocity? Assume that "North" is +y and "East" is +x.
     
  16. Sep 8, 2009 #15
    would you use 9.12 m/s and plug that into the pythagorean theorum to get the initial velocity. 9.12 m/s^2 + 9.12 m/s^2 = 166.35 m/s and take sqr root = 12.9 m/s
     
  17. Sep 8, 2009 #16

    kuruman

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    The Pythagorean Theorem is no help here. If a car is moving North at a speed of 9.12 m/s, what is the northward component of the velocity vector and what is the eastward component of the velocity vector? You need to supply two numbers.

    Northward =

    Eastward =
     
  18. Sep 8, 2009 #17
    I am so lost..I do not understand what you're saying. The only number I can come up with is 9.3 m / 1.60 s = 5.81 m/s. I don't know what other number it could possibly be?
     
  19. Sep 8, 2009 #18

    kuruman

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    We have already established that the speed is

    speed = 14.6 m/1.60 s = 9.12 m/s

    so no more division is necessary. Let me ask you this. Do you understand vectors? Have you studied them? You need to have some understanding of vectors before you can tackle this problem. I will give you part of the answer, but you have to make an effort to understand vectors.

    The initial velocity of the car is given by

    [tex]\vec{v}_{initial}=0\ \hat{i}+9.12\ \hat{j}[/tex]

    The above expression says that initially, the car is moving with 0 m/s in the eastward direction and 9.12 m/s in the northward direction. In other words, the car is moving due north at 9.12 m/s.

    Now can you write the final velocity vector in a similar fashion?
     
  20. Sep 8, 2009 #19
    Yes, I have studied vectors before in Biomechanics..I understand them, I just get confused by physics in general. So you're saying that initial velocity is 0 m/s. So the change in velocity would be 9.12 m/s?
     
  21. Sep 8, 2009 #20

    kuruman

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    I never said that the initial velocity is zero. Look at the equation I gave you. Does it say

    [tex]\vec{v}_{initial}=0 \ ?[/tex]

    No. What does it say instead? Hint: I told you what is says in my previous posting.
     
    Last edited: Sep 8, 2009
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