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At t=0 , a particle starts at the origin with a velocity of 6 feet per second and moves along the x-axis in such a way that at time t its acceleration is 12t^2 feet per second per second. Through how many feet does the particle move during the first 2 seconds?

  • calculus -

    a = 12t^2
    velocity = 4t^3 + c
    when t = 0, v = 6
    6 = 4(0) + c
    c = 6
    v = 4t^3 + 6
    distance = t^4 + 6t + k
    when t = 0, distance = 0 ---> k = 0

    distance = t^4 + 6t
    so when t = 2
    distance = 2^4 + 6(2) = 28 feet

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