calculus

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a particle moves along a number line measured in cm so that its position at time t sec is given by s=72/(t+2) +k, k is a constant and t>=0 seconds.
(a) Find the instantaneous velocity of the particle at t=4 seconds
(b) Find the acceleration of the particle when t =4 seconds
(c) If we know the particle is at s=20 when t=4sec, use your answer from part( a) to approximate the position of the particle at t=4.5 sec.

  • calculus -

    a)v(4)=-2 seconds
    b)a(4).67

  • calculus -

    i don't know the answer for partc.

  • calculus -

    is answer a and b correct

  • calculus -

    a) and b) are correct

    for c)
    sub in the given values
    20 = 72/6 + k
    20 = 12 + k
    k = 8

    so now you know
    s= 72/(t+2) + 8
    when t = 4/5
    s = 72/4.5 + 8 = 24

  • calculus -

    Part (a) is correct, and (b) is correct to 2 decimal places.

    For part (c), you can make use of the approximation formula
    f(t0+h)=f(t0)+h*f'(t0) approximately, and when h is relatively small.
    t0=4
    f(t0)=20,
    f'(t0)=-2
    h=4.5-t0=0.5
    f(t0+h)=f(4.5)=20+(-2)*0.5=19 (approx.)

    Check by initial conditions:
    f(4)=20
    substitute in s(t) to get
    20=72/(4+2)+k=12+k
    => k=8
    So
    f(t)=72/(t+2)+8
    f(4.5)=72/(4.5+2)+8
    =19.077...
    So approximation above is reasonably accurate.

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