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Velocity and acceleration question

  1. Sep 18, 2003 #1
    Hello, I am having trouble deciphering the following problem:

    You start from rest at point i: First, moving at 30 degrees north of due east:

    1. increase your speed at 0.400 m/s^2 for 6.00s

    okay, with this information i did:

    V = Vo + at --> V = 0 + (0.400)(6.00) = 2.40 m/s
    V = d/t --> 2.40 = d/6.00 --> d = 14.4 m

    2. with whatever speed you then have, move for 8.00 s

    Okay, that "whatever speed" I found to be 2.40 m/s. Moving along for 8.00 s, I was thinking that I could do:

    V = d/t --> 2.40 = d/ 8.00 -- > d = 19.2 m

    3. then slow at 0.400 m/s^2 for 6.00s.

    Okay, here is where I get confused. I thought that in parts 1 AND 2, you were still going at the same speed and thus same acceleration. So, why or how could you SLOW at 0.400 m/s^2 when that was the a given in 1?

    Could someone explain to me what exactly is going on? thanks!
  2. jcsd
  3. Sep 18, 2003 #2
    There is an error here. You found the final velocity, which is half the problem. The answer did ask you for the final velocity, but you applied it the wrong way in the second step. To find the distance traveled you must use average velocity. The object did not travel at 2.4 m/s for 6 seconds, he accelerated at 0.4 m/s^2 for 6 seconds (i.e. he was not traveling at 2.4 m/s the whole time). v is the average velocity:
    v = 0.5(vf+v0)
    I'll let you solve it.

    The second step is fine.

    I think what is confusing you is your definition of acceleration (which is wrong). acceleration is how much your velocity changes. when you are accelerating at 0.4 m/s2 that means you are increasing your velocity by 0.4 m/s every second. if you are traveling at a constant velocity (i.e. 2.4 m/s) than your acceleration is 0 m/s2. In this problem the object accelerates .4 m/s every second for 6 seconds. Then it stays at that constant velocity for eight seconds (no acceleration) and then it begins to accelerate at -0.4 m/s2 (or if you like deaccelerate at 0.4 m/s2)

    #EDIT: *Sigh* i'm such an idiot. i accidentally added a t in the equation for average velocity (which would make it an equation for distance (s)). It's now corrected.

    Last edited: Sep 18, 2003
  4. Sep 20, 2003 #3
    Okay, thanks much! I just wanted to check if so far I'm doing this right. So, because it is deaccelerating in part 3, can I do the following:

    X-Xo = Vot + 0.5at^2 = (2.40)(6.00) + 0.5(-0.400)(6.00)^2 = 7.20 m

    But then, if I do:

    V = Vo + at
    V = 2.40 + (-0.400)(6.00) = 0

    So how can that be that there is no velocity at that point?
  5. Sep 20, 2003 #4


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    Staff Emeritus
    Science Advisor
    Gold Member

    Okay. If the object starts out with zero velocity, gains velocity at a given rate for a given time, and then later loses velocity at the same rate for the same amount of time, then of course it is going to have zero velocity at the end.
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