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Kinematics, 2 questions

  1. Apr 17, 2012 #1
    Question 7
    a) Acceleration is the rate of change of velocity

    Use your definition in (a) to show that v = u + at, where v is the final velocity, u is the initial velocity and a and t are the acceleration and the time interval respectively.

    State the conditions that must be satisfied for the equation to be valid.


    Question 11
    An aircraft has a landing velocity of 50m/s and decelerates uniformly at 10m/s^2 until its velocity is reduced to 10m/s. Calculate
    a) The time taken to slow down to this velocity, (answer is 4s but how do I get it?)
    b) The distance covered during the deceleration. (answer is 120m but how do I get it?)

    Need help with these two. Sorry for the mess.

    Thanks guys!
     
  2. jcsd
  3. Apr 17, 2012 #2
    7a) If acceleration is the rate of change of velocity, how could you express a in terms of v, u and t? Rearrange for v

    11a) Using the equation given in 7a, you need to pick out the numbers given in the question and plug them into the correct places to get out the answer you want. Can you do this?
    b) You're going to need another equation involving distance. Do you know one or are you expected to derive it?
     
  4. Apr 17, 2012 #3
    Got the first part of 7, thanks. Don't understand when they ask to "State the conditions that must be satisfied for the equation to be valid." though :/

    For 11, what I've done so far is
    v = u + at
    10 = 50 + 9.81(t)
    -40 = 9.81(t)
    t = -4.0s

    It's a negative number, I guess I'm supposed to flip it around?

    As for part (b), I'm not sure myself. The equations I have around the paper are
    v = u + at
    v^2 = u^2 + 2as
    s = ut + 1/2 at^2

    I've tried using the s = ut + 1/2 at^2 one but I got an answer of 278.48m which is apparently wrong.

    Thanks.
     
  5. Apr 17, 2012 #4
    Good to hear!

    The wonderful thing about maths is that there should be no need to "flip it around" the maths should do it all for you and you have made a mistake. Keep in mind that the plane decelerates at 10ms^-2. Can you think what might cause your stray negative sign to disappear now?

    Given what I've said above, you've made the same mistake in the second part, so you should now be able to fix both.

    I'll also just remind you that the question says the acceleration is 10ms^-2, but your second answer suggests that you are using 9.81, make sure you are being consistent (the answer quoted in your first post [120m] is the correct answer if you use a deceleration of 10ms^-2)
     
  6. Apr 17, 2012 #5
    YES! Got it all! Homework completely, I can rest easy now. Thanks so much for all your help! :D
     
  7. Apr 17, 2012 #6
    Excellent, no problem
     
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