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

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

9.000

-9.000
10.596
-3.00

Those are the answer choices for #2

Let's go through each question step by step:

2. The displacement can be found by integrating the velocity function over the given time interval. The integral of v(t) = -t^2 + 6 is given by the formula:

∫(-t^2 + 6) dt

Integrating with respect to t gives us:

- (1/3)t^3 + 6t

Now we can find the displacement by evaluating the expression at the limits of integration:

Displacement = [- (1/3)(3^3) + 6(3)] - [- (1/3)(0^3) + 6(0)]
= [-27 + 18] - [0 + 0]
= -9 feet

Therefore, the displacement of the object on the interval t = 0 to t = 3 seconds is -9 feet.

3. To find the velocity, we need to integrate the given acceleration function, a(t) = cos(t) - sin(t), with respect to t. The initial velocity v(0) = 3 helps us determine the constant of integration.

Integrating a(t) gives us:

∫(cos(t) - sin(t)) dt = sin(t) + cos(t) + C

We can determine the value of C by using the initial condition, v(0) = 3:

sin(0) + cos(0) + C = 3
1 + 1 + C = 3
2 + C = 3
C = 1

Hence, the velocity function is v(t) = sin(t) + cos(t) + 1.

4. The time it takes for the ball to reach its highest point can be found by considering that at its highest point, the velocity of the ball will be zero.

The velocity of the ball at any given time t is given by the formula:

v(t) = initial velocity + (acceleration due to gravity) * t

Plugging in the given values, we have:

0 = 72 + (-32) * t

Solving for t gives us:

32t = 72
t = 72/32
t = 2.25

Therefore, it will take the ball 2.25 seconds to reach its highest point.

To summarize:
2. The displacement is -9 feet.
3. The velocity function is v(t) = sin(t) + cos(t) + 1.
4. It will take the ball 2.25 seconds to reach its highest point.

To find the solution for each of these questions, we will go step by step.

1. For question #2:
To find the displacement of the object on the interval from t=0 to t=3, we need to calculate the definite integral of the velocity function from t=0 to t=3. The displacement is the area under the velocity curve on this interval.

Given: v(t) = -t^2 + 6

To find the displacement, we integrate the velocity function:
s(t) = ∫[0 to 3] v(t) dt

s(t) = ∫[0 to 3] (-t^2 + 6) dt
s(t) = [-t^3/3 + 6t] evaluated from 0 to 3

Evaluating the definite integral:
s(t) = [-(3^3)/3 + 6(3)] - [-(0^3)/3 + 6(0)]
s(t) = -27/3 + 18 - 0
s(t) = -9 + 18
s(t) = 9 feet

Therefore, the displacement of the object on the interval t=0 to t=3 seconds is 9 feet.

2. For question #3:
We need to find the velocity, v(t), given the acceleration function, a(t), and the initial velocity, v(0).

Given: a(t) = cos(t) - sin(t) and v(0) = 3

To find the velocity, we need to integrate the acceleration function:
v(t) = ∫ a(t) dt

v(t) = ∫ (cos(t) - sin(t)) dt
v(t) = sin(t) + cos(t) + C, where C is the constant of integration

To find the value of C, we use the initial velocity condition:
v(0) = 3

Plugging in t=0 and v(0)=3 into the velocity equation:
3 = sin(0) + cos(0) + C
3 = 0 + 1 + C
3 = 1 + C
C = 2

Therefore, the expression for the velocity, v(t), is:
v(t) = sin(t) + cos(t) + 2

3. For question #4:
To find the time it takes for the ball to reach its highest point after leaving the pitcher's hand, we need to find the time when the velocity becomes zero.

Given: initial velocity = 72 ft/sec and acceleration due to gravity = -32 ft/sec^2

The velocity as a function of time can be found using the equation of motion:
v(t) = v0 + at

Since the acceleration due to gravity is acting opposite to the direction of the initial velocity, the velocity will decrease until it becomes zero at the highest point.

Setting v(t) = 0 and solving for t:
0 = 72 - 32t
32t = 72
t = 72/32
t = 2.25 seconds

Therefore, it will take the ball 2.25 seconds to reach its highest point after leaving the pitcher's hand.

In summary:
- The displacement in question #2 is 9 feet.
- The velocity in question #3 is v(t) = sin(t) + cos(t) + 2.
- The time it takes to reach the highest point in question #4 is 2.25 seconds.