Question

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

Answer 1

9.000
-9.000
10.596
-3.00

Those are the answer choices for #2

Answer 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