Post a New Question

ap calc

posted by .

a particle moves along the x-axis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie-5)t-sin(piet)

a.find the acceleration at any time t.

b.find the minimum acceleration of the particle over the interval [0,3]

c.find the maximum velocity of the particle over the interval [0,2]

  • ap calc -

    I will read that as
    v(t) = (2π - 5)t - sin(πt)
    = 2πt - 5t - sin(πt)

    a(t) = 2π - 5 - πcos(πt)
    for a min of a , a ' (t) = 0

    π^2 sin(πt) = 0
    sin(πt) = 0
    πt = 0 or πt = π or πt = 2π
    t = 0 , t = 1 , t = 2
    a(0) = 2π - 5 - πcos0 = π - 5
    a(1) = 2π - 5 - πcosπ = 2π-5+π = 3π-5
    a(2) = 2π - 5 - πcos2π = 2π - 5 - π = π - 5
    a(3) = ... = 3π - 5

    looks like the min is π-5

    c) for max velocity, a(t) = 0
    2πt - 5t - sin(πt) = 0
    sin (πt) = 2πt - 5t
    very messy equation to solve,
    I ran it through Wolfram's "magic" webpage and got
    t = 0 and t = .670349

    so
    v(.670349) = 0
    v(0) = sin0 = 0
    v(2) = (2π-5)(2) - sin 2π
    - 2.566.. - 0 = 2.566

    check my arithmetic

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question