AP Calculus AB

2. For an object whose velocity in ft/sec is given by v(t) = -t^2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 3 secs?

3. Find the velocity, v(t), for an object moving along the x-axis if the acceleration, a(t), is a(t) = cos(t) - sin(t) and v(0) = 3.

v(t) = sin(t) + cos(t) + 3
v(t) = sin(t) + cos(t) + 2
v(t) = sin(t) - cos(t) + 3
v(t) = sin(t) - cos(t) + 4

4. A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is -32 ft/sec2, how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.

2.25
2.5
4.25
4.5

I'm 90% sure that #2 is -3.00, #3 and #4 im lost. I think #4 is 2.25 but don't know #3

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  1. 9.000
    -9.000
    10.596
    -3.00

    Those are the answer choices for #2

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  2. The concept you are dealing with has to do with distance (s), velocity (v), and acceleration (a)
    You hopefully learned that
    ds/dt = velocity
    and dv/dt = acceleration

    in #2
    v(t) = -2t + 6
    then s(t) = -t^2 + 6t + c
    so s(3) = -9 + 18 + c = 9+c
    and s(0) = 0+0+c
    displacement = 9+c - c
    = 9

    #3
    a(t) = cost - sint
    then v(t) = sint + cost + k
    v(0) = sin0 + cos0 + k = 3
    0+1+k=3
    k = 2

    v(t) = sint + cost + 2 , which is choice #2

    #4
    you should be able to find
    s(t) = -16t^2 + 72t as the distance equation.
    the vertex of this downwards parabola gives you the answer.
    s(t) = -16t(t - 4.5)
    the x-intercepts are 0 and 4.5, with the vertex half-way between them or t = 2.25 seconds
    Your were right on that one

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