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Math Help - A particle moves according to a law of motion (Derivatives Chapter)

  1. #1
    s3a
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    A particle moves according to a law of motion (Derivatives Chapter)

    I don't understand exactly what the question is asking for in (a) and for (e), I feel like I know what it's asking but I'm getting 32 feet which is exactly a third of the correct answer if that helps.

    Question
    :

    43.

    A particle moves according to a law of motion s = f(t), t >= 0, where t is measured in seconds and s in feet.
    (a) Find the velocity at time t.
    (b) What is the velocity after 3s?
    (c) When is the particle at rest?
    (d) When is the particle moving in the positive direction?
    (e) Find the total distance traveled during the first 8s.
    (f) Draw a diagram like Figure 7 to illustrate the motion of the particle.


    Answer:
    (a)3t^2 - 24t + 36
    (e) 96 feet

    My work is attached (#43).

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    It is really hard to help you when we dont know what f(t) stands for.
    I am going to assume that f(t) is a function of position with respect to time.

    <br />
\frac{df}{dt}=3t^2-24t+36<br />

    Is the function of velocity at time t.
    I guess that you put t=0, and got 36 feet/sec, but this is now what you are suppose
    to do. \frac{df}{dt} is the function of velocity at time t.

    For e)
    Notice that when the particel is at rest (t=2) it is turning back, and stops again at t=6.
    Then it turns around for the 2nd time, and goes back to s=32 (at t=8).
    That means that the particle goes forward 32 feet then backwards 32 feet, and forward
    again for 32 feet. So what is the total distance?

    Here is plot of f(t)



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  3. #3
    MHF Contributor Calculus26's Avatar
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    For a) you got the right answer but didn't recognize you had it

    they are asking for v(t) = 3*t^2 -24 t +36 which you had calculated correctly

    For e it is a little more complicated
    between t =0 and t= 2 the velocity is positive so the distance is:

    s(2) - s(0) = 8 - 48 +72 = 32

    between t= 2 and t= 6 the velocity is negative so the distance is

    -[s(6) -s(2)] = 0 + 32

    Between t =6 and t = 8 the velocity is positive again so the distance is

    s(8) -s(6) = 32

    Adding them all up you get 96.

    It is only if the velocity is posive over the interval a to b is the diatance

    s(b)-s(a)
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  4. #4
    MHF Contributor Calculus26's Avatar
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    make sure you read my edited post-- I had writtten the wrong thing in one place
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