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Jet accelerating in problem with Constant Acceleration

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A jet plane is cruising at 300 m/s when suddenly the pilot turns the engine up to full throttle. After travelling 4km, the jet is moving 400 m/s. What is the jet's acceleration?

    known:
    x0 = 0m, t0 = 0s, v0 = 300 m/s
    x1 = 4000m, t1 = ?, v1 = 400 m/s


    2. Relevant equations
    t1 = x1 - x0 / v1 - v0
    a = v1 - v0 / t1


    3. The attempt at a solution
    t1 = x1 - x0 / v1 - v0 = 4000m / 100 m/s = 40s
    a = v1 - v0 / t1 = 100m/s / 40s = 2.5 m/s/s

    The book says the acceleration is 8.8m/s/s. I have worked this problem several ways and cannot seem to come up with the solution
     
  2. jcsd
  3. May 5, 2014 #2

    Doc Al

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    Staff: Mentor

    This reasoning is not correct. To find the time it took to cover those 4 km you need the average speed during that interval. Which is what? (For some reason, you used the change in speed.)
     
  4. May 5, 2014 #3
    In solving for t1 you incorrectly assumed that the velocity was a constant 100 m/s.

    Do you another kinematic equation for distance under a constant acceleration?
     
  5. May 5, 2014 #4
    I assumed it was 100m/s because it starts off at 300 m/s and ends up at 400 m/s after 4000m and so the difference is 100 m/s. the other ones I have is the standard one:

    x1 = x0 + v0(t1 - t0) + 0.5 * a(t1 - t0)^2. Substituting in the values I come up with:

    4000 = 300(40) + 0.5 * a(40)^ = 4000 = 12000 + 0.5 * a * 1600
    -8000 = 800*a
    10 m/s/s = a.

    Or I could use this one:
    v1^2 = v0^2 + 2a * (x1 - x0)
    400 ^ 2 = 300 ^ 2 + 2a * 4000)
    160000 - 90000 = 8000a
    70000 = 8000a

    ...

    Okay I swear I did this several times.... grrrr.

    Thanks!
     
  6. May 5, 2014 #5
    So my question now is how do I know which of the three kinematic equations to use? All three provide through algebraic methods the ability to find the acceleration? But I have gotten different answers from all three. The ciorrect answer being in the third one.
     
  7. May 5, 2014 #6

    Doc Al

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    Staff: Mentor

    That's certainly the difference in speed, but that's not what you need to calculate the time. To calculate the time, you can use distance = ave speed * time, but first you need the average speed.

    This is wrong because you are assuming a time of 40 s, which is wrong.

    That's fine.
     
  8. May 5, 2014 #7

    Doc Al

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    Staff: Mentor

    Your only mistake was using this equation, which is not valid:
     
  9. May 5, 2014 #8
    I recalculated it using the average and yes I was able to get 8.8 m/s^2 again. So is it safe to say that whenever you're given two velocities and asked to find something between that you should find the average and calc based on that?
     
  10. May 5, 2014 #9
    Technically, these kinematics equations only work when you know the acceleration is constant. But for practical purposes, virtually all kinematics questions assume constant acceleration, so yes.
     
  11. May 5, 2014 #10
    awesome thank you!!!
     
  12. May 5, 2014 #11

    Doc Al

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    Staff: Mentor

    Well, it depends on what you need to calculate and how you plan to do it.

    If you want to use a kinematic formula that involves the time, then you'll need to calculate the time. For that you can use Δx = vave*Δt.

    A useful list of kinematic formulas can be found here: Basic Equations of 1-D Kinematics
     
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