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Homework Help: Object undergoing acceleration

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    An object undergoes acceleration 2.3i + 3.6j for 10s. At the end of this time, its velocity is 33i + 15j.
    a) What was its velocity at the beginning of the 10s interval?
    b) By how much did it's speed change?
    c) By how much did its direction change?
    d) Show that the speed change is not given by the magnitude of the acceleration multiplied by time. why?

    3. The attempt at a solution

    a) a→= Δv/Δt

    vi→ = (33i + 15j) - (23i + 36j)
    vi→ = (10i -21j) ms^-1

    b) |vi→| = 23.26ms^-1
    |vf→| = 36.25ms^-1
    speed = |v| = (|vi→|) -|vf→|
    |v| = 13ms^-1

    c)tanΘ = 15/33

    d) |a→| = sqrt (2.3^2 + 3.6^2) = 4.272ms^-1
    Δt = 10s

    |a→|.10s [itex]\neqΔv[/itex]
    ∴4.272ms^-1 . 10s [itex]\neq13ms^-1[/itex]

    what is the explanation?
  2. jcsd
  3. Jan 18, 2014 #2
    what do you know about the initial directions of the velocity and acceleration?
  4. Jan 18, 2014 #3
    Could you be more specific?
  5. Jan 18, 2014 #4


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    How are velocity and speed defined? Is it possible that the velocity changes and the speed does not?
    If the velocity is [itex] \vec {v } = \vec {v_0} +\vec {a } t [/itex] what is the change of the speed during time t?

  6. Jan 18, 2014 #5
    Is this question with respect to part (d)?
  7. Jan 18, 2014 #6


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    yes, of course.

  8. Jan 18, 2014 #7
    Are my reasoning and workings valid from part(a) - part(c)?

    In respond to your earlier question, the change in velocity is a function of time.
    vf = vi + at

    a.t = Δv
    a = (2.3i + 3.6j)

    Δv = 10*(2.3i + 3.6j) = 23i + 36j
  9. Jan 18, 2014 #8


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    It was the change of speed.
    That is the angle of the final velocity Θf. What is the initial angle? By how much did the direction change?

    The change of velocity is

    [tex]\vec v_f - \vec v_i=\vec a t [/tex]

    You can make a triangle with side-lengths vf,vi and at. What do you know about the sides of a triangle? Can you draw a triangle with one side equal to the difference of the two other sides? How does it look like? What relation is true among the side lengths of a general triangle?
  10. Jan 18, 2014 #9
    final angle = 24.4 degrees
    initial angle = arctan (-21/10) = -64.5 degrees
    Θchange = Θf - Θi = 89°

    But isn't the change in velocity mathematically reasoned to be a.t?

    dv/dt = a
    a.dt = dv

    Edit: it's the same thing.

    In general, the sides can be expressed as ratio of one another. This property effective allows us to determine the angle between the lengths.
    Last edited: Jan 18, 2014
  11. Jan 18, 2014 #10


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    Yes, the change of velocity (vector) is the acceleration (vector) multiplied by time. But that is not true to the change of speed.
    Look at your triangle. The length of one side is equal to the initial speed |v1|, the other is equal to the final speed |v2|, and the length of the third side is |a|t - magnitude of acceleration multiplied by time. Is |v2|-|v1|=|a|t?

  12. Jan 18, 2014 #11
    Yes I noticed.

    vf - vi = a.t [itex]\neq sqrt[vf^2 + vi^2][/itex]

    Edit: yes it.
  13. Jan 18, 2014 #12


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    You need to write some explanation which holds for any vector difference. In this case, the initial and final velocities are almost perpendicular. Compare the magnitude of the difference and the difference of the magnitudes in general case.

  14. Jan 18, 2014 #13
    I'll contemplate on the reasoning symbolically.
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