Physics

Question

A particle moves on a straight line with velocity function v(t) = sin ωt cos2ωt. find its position function s = f(t) if f(0) = 0.

2 Answer

  • The correct answer for this question is this one:Hope this helps answer your question and have a nice day ahead.
    v( t)= sin( ωt) cos^2 ( ωt) 

    s(t) = ∫sin(ωt) cos^2 (ωt) dt 

    let cos(ωt) = u 

    -ωsin(ωt) dt = du 

    ∫sin(ωt) cos^2 (ωt) dt = -(1/ω)∫u^2 du = -(1/ω)u^3 / 3 + const 

    s(t) = -(1/ω)cos^3(ωt) / 3 + const 

    s(0) = 0 => const = 1/(3ω) 

    s(t) = (1 - cos^3(ωt))/(3ω)
  • By integrating the velocity equation, we will see that the position equation is:

    [tex]f(t) = \frac{cos^3(\omega t) - 1}{3}[/tex]

    How to get the position equation of the particle?

    Here we know that the velocity of the particle is:

    [tex]v(t) = sin(\omega t)*cos^2(\omega t)[/tex]

    To get the position equation we just need to integrate the above equation:

    [tex]f(t) = \int\limits {sin(\omega t)*cos^2(\omega t)} \, dt[/tex]

    If we take:

    u = cos(ωt)

    Then:

    du = -sin(ωt)dt  ⇒  dt = -du/sin(ωt)

    Replacing that in our integral we get:

    [tex]\int\limits {sin(\omega t)*cos^2(\omega t)} \, dt\\\\-\int\limits {\frac{sin(\omega t)*u^2 \, du}{sin(\omega t)}\\\\\\-\int\limits {u^2} \, dt = -\frac{u^3}{3} + c[/tex]

    Where C is a constant of integration.

    Now we remember that u = cos(ωt)

    Then we have:

    [tex]f(t) = \frac{cos^3(\omega t)}{3} + C[/tex]

    To find the value of C, we use the fact that f(0) = 0.

    [tex]f(t) = \frac{cos^3(\omega *0)}{3} + C = \frac{1}{3} + C = 0\\\\C = -1/3[/tex]

    Then the position function is:

    [tex]f(t) = \frac{cos^3(\omega t) - 1}{3}[/tex]

    If you want to learn more about motion equations, you can read:

    https://brainly.com/question/19365526

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