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I got stuckplease

  1. Sep 4, 2005 #1
    A car with a velocity of 25m/s [E] changes its velocity to 25m/s in 15s .
    Calculate the car's average acceleration.

    I have found out that the direction of the acceleration is 45degrees s of w.
    But I have no idea about the magnitude of it...


    and another similar problem is that:

    a watercraft with an initial velocity of 0.4m/s [E] undergoes an averager acceleration of 2 m/s/s for 2.5s. waht is the final velocity of ther watercraft?


    Thank you very much...
     
  2. jcsd
  3. Sep 4, 2005 #2

    Curious3141

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    Work completely in terms of vectors. I prefer to use complex numbers to represent them, because I find them more natural. Have you learned complex numbers and their representation on an Argand diagram ?

    If you don't know how to do that, you can just use unit vectors [tex]\vec{i}[/tex] and [tex]\vec{j}[/tex] on a 2-d plane.

    Always go back to the definition of acceleration as change in velocity divided by time interval. After you get the vector expression for the acceleration, take the magnitude and direction of that.
     
  4. Sep 4, 2005 #3
    I am not sure about the first question. Was there more information given, like the initial velocity of the car?

    As for the second question use the formula acceleration=(change in velocity)/time
     
  5. Sep 4, 2005 #4

    Curious3141

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    As an example of how you would use complex numbers in the first question, let velocities along the E-W axis be purely real, and those along N-S be purely imaginary. An arbitrary velocity vector would be a complex number (sum of real and imaginary components).

    Initial velocity [tex]v_i = 25[/tex]

    Final velocity [tex]v_f = -25i[/tex]

    Hence change in velocity [tex]v_f - v_i[/tex] = ?

    Time interval = [tex]\Delta t[/tex] = 15 seconds.

    Hence acceleration = [tex]a = \frac{v_f - v_i}{\Delta t}[/tex] = ?

    Magnitude of acceleration = [tex]|a|[/tex] = ?

    Direction of acceleration = [tex]arg(a)[/tex] = ?

    Can you fill in the blanks ?
     
    Last edited: Sep 4, 2005
  6. Sep 4, 2005 #5

    VietDao29

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    He probably hasn't learned Complex number, Curious3141.
    By the way, there's no need to make the problem more complicated. The normal 2-d plane works well.
    Viet Dao,
     
  7. Sep 4, 2005 #6

    Curious3141

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    Right you are. :smile: I just find this technique more "natural", since I learnt complex numbers (on my own) well before I learnt what a vector was (in school).
     
  8. Sep 4, 2005 #7

    HallsofIvy

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    Finding the magnitude should be easier than finding the angle! Of course, acceleration is change in velocity divided by change in time. You have correctly seen that change in velocity is the difference of the two vectors. One way to do that is to draw the two "arrows", tail to tail, and then draw the "arrow" from the tip of the E arrow to the tip of the S arrow. That "arrow" is the change in velocity vector. You observed that this is an iscosceles right triangle and so has angles of 45 degrees. Yes, the change of velocity "arrow" is pointing 45 degrees S of W. Now, since this is a right triangle, find the magnitude of that change by using the Pythagorean theorem. Finally, of course, divide that by 15s to find the magnitude of the acceleration vector (the angle doesn't depend on the time).

    What some of these other people are trying to say (in their own inimitable manner) is that, unless it is required, you don't actually have to deal with "angles" and "magnitude". Set up a coordinate system so that the positive x-axis is East and the positive y-axis is N. Then the initial velocity vector is 25i+ 0j and the final velocity vector is 0i-25j. Subtracting those 2 vectors (final minus initial) , the change of velocity is -25i- 25j. Since that occured in 15 seconds and acceleration= change in velocity/change in time, the acceleration vector is -25/15i- 25/15j= -5/3i- 5/3j m/s2. If you calculate the angle and magnitude of that, you will get the same as before.


    acceleration time time is change in velocity: the change in velocity is 2*2.5= 5.0 S. Draw the initial velocity vector of "length" 0.4 pointing E and, from the "point" of that, the change in velocity of "length" 5 pointing S. You should see that you have a right triangle with legs of length 0.4 and 5.0. You can find the magnitude of the sum (the hypotenuse) by using the Pythagorean theorem and the angle by using trig functions.

    Or, using the "coordinate" method I mentioned above, write the initial velocity vector as 0.4i+ 0j and the change of velocity as 0i- 5.0j and add: the final velocity is 0.4i- 5.0j.


    You're welcome
     
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