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Distance traveled correct?(not homework)

  1. Jan 8, 2009 #1
    I am currently doing the self-test in Ch1 of Basic Physics: A Self-Teaching guide, by Karl F. Kuhn. I am trying to teach myself the basics of physics so I can be ready for March's session of physics class. I had a go at it for 2 days and realized i was going to fail unless I did something to be more comfortable with it.

    ok Question 3 says:

    3) an object traveling with the speed of 2 m/s. Three seconds later its speed is 14 m/s. How much is its acceleration during this time. Answer= 4 m/s

    I get this by using the acceleration equation of a = (V1-V0)/t or (14-2)/3= 4 m/s

    now my issue is question 5 which says:

    5)How far does the object in question 3 travel during the 3 seconds of its accelerated motion?

    so I worked it like this using the acceleration, distance traveled and time of fall relation equation.

    d = V0(t)+1/2 * at^2 or 2(1) +1/2(4.67)(3)^2 = 23.02 m

    for the V0(t) i used 2m/s times the assumed time of speed after 1 second

    for the a (or acceleration) i averaged over 3 seconds the total speed of 14 m/s or 4.67 m/s

    the obviously, following the equation, multiplied by the the total time of 3s squared

    the book is trying to tell me 24 m is the answer, can someone tell me what I did wrong here, or is the book incorrect?
  2. jcsd
  3. Jan 8, 2009 #2

    Doc Al

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    Staff: Mentor

    In the kinematic equation for distance, t is the time. Why did you put t = 1 in the first term but t = 3 in the second term?? It's given that t = 3.

    I have no idea what you're doing here. You already calculated the acceleration to be 4 m/s^2 (not m/s, by the way).

    Fix these errors and redo.
  4. Jan 8, 2009 #3
    Ok I found my mistake. I averaged the total speed by mistake and used it in place of the acceleration.

    here is the correct equation set up.

    2(3) +1/2(4)(3)^2=24

    I must think too far into the problem, and gotta slow down.
    Last edited: Jan 8, 2009
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