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Homework Help: Setting intantaneous velocity equal to average velocity

  1. Sep 3, 2010 #1
    1. an object moving at constant acceleration. Find the average velocity between the times t1 and t2, with t2 > t1. Find the time t∗ between t1 and t2 such that
    the instantaneous velocity at t∗ equals the average velocity between t1 and t2.




    2. [X(t) = X(o) + V(o)t + 1/2(at^2)], [v0 + a(t2 + t1)/2], [v0 + a t1]



    3. the first thing i did was find the average velocity betwee the times t1 and t2. this portion was just given as definition in my book. my book provides this conversion.

    {[X(o) + V(o)t2 + 1/2(at2^2)] - [X(o) + V(o)t1 + 1/2(at1^2)]}/(t2-t1), they then go on to state a simplification of the above formula as [v0 + a(t2 + t1)/2]


    this seems like the average velocity they want, t2-t1 so that velocity will remain positive is t2 is greater. the second part seems less clear to me.

    then they as me to Find the time t∗ between t1 and t2 such that
    the instantaneous velocity at t∗ equals the average velocity between t1 and t2. the book gives me the equation [v0 + a t1](i suppose the derivative of constant acceleration) for instantaneous velocity and im not sure what to do. am i supposed treat t1 and t2 in the average velocity equation as one variable, solve for that t, and then observe it equal to instantaneous rate solved for its t? i am really at a loss here, any help would be appreciated.
     
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  3. Sep 3, 2010 #2

    vela

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    For a given t1 and t2, you found the average velocity over this time interval is given by

    [tex]\bar{v} = v_0 + \frac{a(t_1+t_2)}{2}[/tex]

    You want to set this to an expression for v(t), the instantaneous velocity at time t, and then solve for t.

    You may want to verify your relevant equations again. You either have a typo above or you're not remembering the one you need correctly, which might be the cause for your confusion.
     
  4. Sep 3, 2010 #3
    thanks so much for your reply, i posted the above shortly after waking up, so it is less than perfect in organization, but what you posted is very similar, if i understand correctly. so the average velocity is [v(o) + a(t2 + t1)/2] and i set that equal to the instantaneous velocity, [v(o) + at1]. what i was trying to express earlier is my confusion about t1 and t2 in the equation for average velocity and whether i just treat them both as a combination variable while solving for t after setting equal to instantaneous velocity

    so it goes(i think)

    v(o) + at = v(o) +a(t + t)/2

    v(o) + at = v(o) +a(2t)/2

    v(o) +at = v(o) +at

    t=o

    t must =o for instantaneous velocity to be the same as the average velocity, is it right?
     
  5. Sep 3, 2010 #4

    vela

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    No, that's not correct. You should try to check your answer with specific cases to see if it gives a reasonable answer. For instance, say you dropped a ball. It starts from rest and one second later it's moving at 9.8 m/s. Clearly, the average velocity over that one-second interval is not 0 m/s, but the velocity at t=0 is 0 m/s.

    You also might want to think about this simple example to get straight what the formulas mean and what the different variables stand for. Also, try looking through your textbook of graphs of v(t) for constant acceleration. See if you can deduce what the answer should be from the graph, then see if you can get that result from the math.
     
  6. Sep 3, 2010 #5
  7. Sep 4, 2010 #6

    vela

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    Let's forget the graph for now.

    Tell me, what do t1 and t2 stand for?
     
  8. Sep 4, 2010 #7
    it is two time of interest making up an interval. t2 being the later point in an interval and t1 being the earlier point in the interval.

    i was mailed this so..
    a. 70 m/s
    b. 75 m/s
    c. 75 m/s
    d. 85 m/s
     
    Last edited: Sep 4, 2010
  9. Sep 4, 2010 #8

    vela

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    OK, so they define the interval of interest. The formula

    [tex]\bar{v} = v_0 + \frac{a(t_1+t_2)}{2}[/tex]

    then lets you find the average velocity [itex]\bar{v}[/itex] over this interval of time while the equation

    [tex]v(t) = v_0 + at[/tex]

    tells you the instantaneous velocity at time t, where t could be any time, independent of the time interval. When you combine the two,

    [tex]v_0+at = v_0 + \frac{a(t_1+t_2)}{2}[/tex]

    and solve for t, you're asking "When is the instantaneous velocity v(t) equal to the average velocity of the interval t1 to t2?" Does that make sense?
    Sorry, I decided to take a different approach so I deleted that post. Anyway, your answers are off because v(0)=v0 in the equations is the velocity at time t=0. The info provided was v(t=10 s). The variables in the formulas have well defined meanings, so don't just plug a number in because it shows up in the problem. You have to make sure the number provided matches up with the meaning of the variable before you can plug it in. If the problem had said v(0)=15 m/s, your answers would be correct.
     
  10. Sep 4, 2010 #9
    yes, i was on this trail earlier, what i dont understand here is what to solve for. i feel that i should be solving for the t on the left, so that the final answer is an expression of t from the instantaneous velocity equation in terms of the time interval (t1+t2) from the average velocity equation like this:

    v(o) + at = v(o) +a(t1+t2)/2 --> t = (t1 + t2)/2

    think i understand what i did wrong here now, for the two instantaneous velocity problems v(o) should have just been 0 rather than 15. is that right? now i say

    a. 70 m/s
    b. 75 m/s
    c. 60 m/s
    d. 70 m/s

    thanks for your help and patience, i ordered up a physical text on this material from ebay, so hopefully that will help me.
     
    Last edited: Sep 4, 2010
  11. Sep 4, 2010 #10

    vela

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    That answer's correct. Why do you solve for t on the left? Because that's the variable that is the answer to the question "When is the instantaneous velocity...?" The variable t is the one associated with the instantaneous velocity while t1 and t2 are the ones associated with the time interval for the average velocity.

    Earlier, you were replacing t1 and t2 with t, which didn't make sense because the three variables all stand for different things. That's why you were getting that non-sensical answer t=0.
    No, the velocity at time t=0 isn't 0. You have to calculate it based on the velocity at time t=10 s and the acceleration.

    If you consider the question "Why would you think it's 0?," I think you'd realize you just made a guess. (Probably because v(0)=0 is common in other problems.) There's nothing in the information I provided that said v(0)=0. You should have a reason for every equation you write down. When you do a problem, this often involves translating the words into equations. For example, if a problem says the object starts from rest, is initially not moving, is dropped, etc., you can typically interpret those phrases to mean v(0)=0.
     
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