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Math Help - Particle Motion - Beginner

  1. #1
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    Particle Motion - Beginner

    Hi, I'm currently revising and I have a question relating to Particle Motion. The question grows, but to get started I have to show that

    The acceleration, a m/s^2, of the particle, is

    a = sin t + 4

    Given that m = 6 (kg) and F(t) = 6 sin t + 24N
    Words do not express how grateful for any help I would be! I'd like to understand how to do this. Thanks.
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  2. #2
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    Just use Newton's Second Law.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Just use Newton's Second Law.
    How? Thank you.
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  4. #4
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    Well, you've got F(t) = 6 sin t + 24 [N]. Newton's Second Law says that F = m a. Therefore, you can solve for a and plug in your F and m, and the result pops right out. Can you show the steps?
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    Quote Originally Posted by Ackbeet View Post
    Well, you've got F(t) = 6 sin t + 24 [N]. Newton's Second Law says that F = m a. Therefore, you can solve for a and plug in your F and m, and the result pops right out. Can you show the steps?
    so I get

    6 sin t + 24 [N] = 6 x a

    a = 6 sin t + 24[N] / 6

    at t = 0, a = 0.406

    Does that look about right?
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  6. #6
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    Well, all you've shown is the value of the acceleration at t = 0. That doesn't finished the job, at least as stated in the OP. Incidentally, you really need to put parentheses in your expression. What you've actually written is

    a = 6 sin t + (24[N] / 6),

    when you really meant

    a = (6 sin t + 24[N]) / 6.

    To finish the problem, just divide the 6 out term-by-term. You're almost there!
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  7. #7
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    F = M x a

    Therefore a = F/M.

    (6Sint + 24N)/6 = Sin t + 4
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    Quote Originally Posted by LukeEvans View Post
    F = M x a

    Therefore a = F/M.

    (6Sint + 24N)/6 = Sin t + 4
    There you go. Don't forget to tag on the correct units at the end (which are what?)
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  9. #9
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    m/s^2 ?

    I hope lol
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  10. #10
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    Looks good to me.
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  11. #11
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    Marvelous, thanked accordingly good sir!

    Unsurprisingly, I have a further question; assuming the particle starts from rest at any datum point, how do I show that the distance traveled, (x), from the datum point at time t is

    x = 2 t^2 + t - sin t
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  12. #12
    Behold, the power of SARDINES!
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    Quote Originally Posted by LukeEvans View Post
    Marvelous, thanked accordingly good sir!

    Unsurprisingly, I have a further question; assuming the particle starts from rest at any datum point, how do I show that the distance traveled, (x), from the datum point at time t is
    Here is a hint: you know the acceleration, but acceleration is the derivative of velocity so this gives


    \frac{dv}{dt}=\sin(t)+4

    So now take the integral to get the velocity and don't forget that v(0)=0(the particle started from rest.)

    Now do the same thing to find the position! See if you can finish from here.
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  13. #13
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    Quote Originally Posted by TheEmptySet View Post
    Here is a hint: you know the acceleration, but acceleration is the derivative of velocity so this gives


    \frac{dv}{dt}=\sin(t)+4

    So now take the integral to get the velocity and don't forget that v(0)=0(the particle started from rest.)

    Now do the same thing to find the position! See if you can finish from here.
    Forgive me for being confused. How does that look? I have broken the question down, this is it in full:

    (b) Given that initially the particle starts from rest at a particular datum point on the line of motion, show that the distance travelled, x, from the datum point by the particle at time t is,

    x = 2 t^2 + t -sin t

    Hence, show that the time taken for the particle to travel 8 m from the datum point is given by the solution of the equation f(t)=0 where:

    f(t) = 2 t^2 + t - 8 -sin t
    I just need to see it happen step by step. Thanks again.
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  14. #14
    Behold, the power of SARDINES!
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    Quote Originally Posted by LukeEvans View Post
    Forgive me for being confused. How does that look? I have broken the question down, this is it in full:



    I just need to see it happen step by step. Thanks again.
    You just need to integrate both sides of the equation with respect to t

    \int \frac{dv}{dt}dt=\int (\sin(t)+4)dt \implies v(t)=-\cos(t)+4t+c

    Now use the initial condition

    v(0)=0

    to solve for c.

    Now just do the exact same thing except this time velocity is the derivative of position this gives

    \frac{dx}{dt}=

    The answer you got above. See if you can finish from here.
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