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Magnitude of Avg Accel?

  1. Dec 10, 2004 #1
    An airplaine is flying North at 200m/s. It make a gradual turn at constant speed. 20 secs later, it completes the turn and is moving East at 200 m/s. What is the magnitude of its average acceleration while making its turn?

    What does it mean by magnitude of Acceleration?
  2. jcsd
  3. Dec 10, 2004 #2
    Magnitude just means size.

    You can tackle this one using the formula for centripetal acceleration, if you know it.

    Or simpler, just look at the change in the velocity and divide by the time taken. Remember, velocity includes direction as well as speed. Even though in this case the speed remains constant, the direction, and therefore the velocity is changing.
  4. Dec 10, 2004 #3
    i thought v = [tex]\frac{\Delta\nu}{\Delta\chi}[/tex]

    right? change in distance over change in time, but the distance would be a quarter of the circumference, and the change in time is 20 secs. so... 20 times 200?
    Last edited: Dec 10, 2004
  5. Dec 11, 2004 #4
    There's more to velocity than (change in distance) / (change in time)

    Consider a car driving past your house heading north, at 60 mph. It passes your house at 12:00:00 slams on the brakes, skids around a J (U) turn and accelerates back past your house at 60 mph heading south at 12:00:15.

    Now in 15 seconds what distance has the car covered? Zero? Has the car accelerated or decelerated?

    You need to consider velocities as vectors. Have you covered that yet? It normally involves drawing arrows where the length of the arrow represents the speed, and the direction the arrow points in represents the direction. You add and subtract vectors by drawing them nose to tail, and then considering the new vector (arrow) that joins the start and end of the chain.
    Last edited: Dec 11, 2004
  6. Dec 12, 2004 #5
    a(c) = v^2/r

    but it's a curve... so how do you add the vectors? :confused:

    I'm So FRUSTRATED! :grumpy:
    Last edited: Dec 12, 2004
  7. Dec 12, 2004 #6
    You don't have to worry about the path taken. That's why they use 'average' in the question. You only have to consider the starting and finishing velocities, and the time taken.

    The plane has lost 200m/s of northward velocity, so it's the same as gaining 200m/s of southward velocity. At the same time it's gained 200m/s of eastward velocity. We can draw the two changes and add them together like this:

    Code (Text):

    | 200
    V      200
    Now you can draw the vector connecting the start and end of this chain and that gives you the effective velocity change. Acceleration is just change in velocity divided by the time taken.

    Edit: I think this will work, but I'm not totally sure. If I were working out this one, I would calculate the radius of the turn (you know the plane flies a quarter of a circle, that is a distance of [tex]2\pi r / 4[/tex]) in a certain time at a constant given speed, so calculating r is easy.

    And you already know the formula for centripetal acceleration [tex]\frac{v^2}{r}[/tex] so you're laughing.

    It will be interesting to see how the answers from the two methods compare. I think now that the vector adding method might give a slightly smaller answer, as a force acting in that direction would slow the plane down somewhat, and then speed it up again, so it's not what the question asks.

    Sorry if I mislead you.
    Last edited: Dec 12, 2004
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