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Homework Help: Deriving the velocity from a distance equation

  1. Oct 6, 2007 #1


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    1. The problem statement, all variables and given/known data

    x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3

    a) find the expression for the velocity as a function of time.

    b) plot the graphs of the position, velocity and acceleration as accleration vs time for the interval given.

    c) At what time(s) btw t=0 and t= 3.00 is the particle at rest?

    d) for each time calculated in part c) is the acceleration of the particle positive or negative?

    e) calculate the accelration of the particle when the velcocity of the particle is at maximum.

    f) does the particle have uniform acclerated motion? explain.

    2. Relevant equations

    x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3

    3. The attempt at a solution

    a) I need to see if my thinking is correct as well as what I did....

    v(t)= x'(t)= 7.00m + (14.00m/s)t - (6.00m/s^2)t^2

    rearranged:v(t)= -(6.00m/s^2)t^2 + (14.00m/s)t + 7.00m

    b) for plotting the graph of velocity and acceleration vs time it would be the equation for my piece and there would be 2 graphs right?

    c.) yes I found it to be when the graph v is flat at the top of the parabola..I see that when plug into my calculator and that is when t or x = 1.027 to t or x= 1.311 where v(t)= 15.11 (the flat part = when the accleration = to 0)

    d.) If the velocity is 0 wouldn't the acceleration be constant and thus would be positive ?
    For this one I'm not sure ...

    e.) How would I find the maximum velocity? would that be the max point in the graph? I assume that is also when the acceleration of the particle is 0 since that is the highest point in the v vs t graph but also where the acceleration is 0.

    f.) I'm confused about this...wouldnt' the acceleration be 0 when the particle's velocity is max??

    g.) I say that the particle does not have uniform accelerated motion since it first starts out as a slanted line toward the right and it is accelerating at a constant rate then it levels off and eventually slants down so the acceleration goes down.

    Is this alright??
    I'm not sure about d, e, and f

    Thank you =D
  2. jcsd
  3. Oct 6, 2007 #2


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    Is my derivation correct? I was wondering about that negative..does it go into the bracket or stay on the outside?
  4. Oct 6, 2007 #3
    your units should remain the same

    i'm still working out your prob, just replying early.

    a. everything is fine except your units and the negative should be wherever you like :D idk really. doesn't bother me.

    b. yes you will need to plot 2 graphs or you can do them on 1 graph which shows the relationship which is more convenient

    c. that is when velocity is at a max, same thing you said in e.

    d. take the derivative of your velocity function

    e. yes

    f. idk

    g. idk
    Last edited: Oct 6, 2007
  5. Oct 6, 2007 #4


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    for part c), they ask when the object is at rest, hence when the velocity is 0... not when the velocity is at its maximum.
  6. Oct 6, 2007 #5


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    a) Is my derivation that I did correct before?

    b)graphed it.

    c) well looking at the graph assuming my derivation was correct, it goes and hits the 0 line for the equation that I derived at only 1 point from 0-3 sec time.
    t= 2.752 while the distance equation graph shows that distance is constant at t= 2.765

    However if the question states that at which times is it constant...would that mean that there is more than 1 time ?

    d)and e)
    would the acceleration be positive?? I don't know but if the velocity is 0 the acceleration is constant...so since:

    d= 30.645m at 0 velocity point t= 2.7s
    v= 0m/s

    is it a=v/t
    so...a= 0/2.7s= 0 accleration when v= 0m/s

    f) I say it doesn't since the graph shows the velocity increasing then decreasing to zero and if it was constant accelerated motion it would just increase or decrease in a straight line instead of being a parabola shape..

    Is this fine?

    Thank you
  7. Oct 6, 2007 #6


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    What exactly do you mean when you say distance is constant? Do you mean that the derivative of distance is 0?

    The velocity should be 0 at exactly the same time that the derivative of distance is 0(ie distance is at a max or a min)

    2.752 and 2.765 are close enough...

    By constant do you mean the derivative of velocity is 0? well, what does your graph say, is the derivative 0 at more than one point?

    Anyway this has nothing to do with the question about where the object is at rest.

    Just check your acceleration graph... is the acceleration positive or negative on the graph at t = 2.765s?

    BTW if velocity is 0 that does not mean acceleration is constant... Rather if the acceleration is 0... then the derivative of the velocity is 0.

    I don't understand this...

    Why don't you just look at your acceleration graph? Is the acceleration constant. ie: is the acceleration a horizontal straight line? If it is then you have uniform accelerated motion... otherwise you don't.
  8. Oct 6, 2007 #7


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    I just scanned the graph...
    http://img254.imageshack.us/img254/6144/lastscanrl6.th.jpg [Broken]

    well about before...
    I was refering to the distance being constant as being d=30 since when the object doesn't move the distance is the same for that time interval but here it is only 1 point specifically when t= 2.7s

    Me: d)and e)
    would the acceleration be positive?? I don't know but if the velocity is 0 the acceleration is constant...so since:

    Learning Physics: (response to above bolded statement)
    ~checking my graph now after I graphed the acceleration...
    I took the derivative of the distance equation given and got...

    x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3

    v(t)= x'(t)= -(6.00m/s^2)t^2 + (14.00m/s)t + 7.00m

    derived again to get accleration...

    a(t)= v'(t)= -(12m/s)t + 14m

    (I find it weird that the acceleration is in m/s but that's what I end up with after derivation of the original equation which involved or did I do the unit derivation wrong? since isn't acclereration always in m/s^2?)

    ~So I get that acceleration is a= -22___ I'm not sure about the units though...

    since that is a negative accleration must be negative right?

    f) based on my acceleration line plot on the graph that is a straight line...I now conclude that it does have uniform acceleration that is negative.

    Is this fine?
    I think so but I don't know how to figure out if the units are allright.

    Thank you :D
    Last edited by a moderator: May 3, 2017
  9. Oct 8, 2007 #8


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    Hi christina. When you take the derivatives:

    x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3

    remember... the units in the coefficients don't change when you take the derivative... they are just constants... leave the units the same.

    v(t) = 7.00m/s + (14.00m/s^2)t - (6.00m/s^3)t^2

    so see the units here will all be m/s when t is plugged in (t is in seconds).

    BTW for part c), you want when v(t) = 0... you can solve this equation to get an exact answer and see if it fits what you got from your graphs...


    a(t) = 14.00m/s^2 - (12.00m/s^3)t

    so these units are m/s^2, once t in seconds is plugged in...

    so acceleration is definitely negative around the time 2.765s... so that's part d)...

    for part e)... the velocity is at maximum or minimum when its derivative = 0... ie when acceleration = 0... what time do you get when you solve for a(t) = 0... this should fit with the time you see on the graph...

    for part f)... no this is not uniform acceleration... uniform acceleration is when acceleration is constant (acceleration is a horizontal straight line)... here acceleration is changing with time.
  10. Oct 8, 2007 #9


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    Thank You learningphysics for your help =)
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