Given two points:

\(\displaystyle A=(x_1,y_1,z_1);B=(x_2,y_2,z_2)\)

The midpoint of \(\displaystyle \overrightarrow{AB}\) is given as:

\(\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})\)

Using our given coordinates, the midpoint is found by simple application of this formula:

\(\displaystyle (\frac{3+2}{2},\frac{5+(-4)}{2},\frac{-10+1}{2})=(\frac{5}{2},\frac{1}{2},\frac{-9}{2})=(2.5,0.5,-4.5)\)

This is one of the answer choices.

To find the position function from the velocity function, take the integral of the velocity function, using the power rule for integration:

\(\displaystyle \int x^n = \frac{x^{n+1}}{n+1}\)

Applying this formula to the given equation, remembering to multiply by the constants in each term:

\(\displaystyle \int12t-6t^2 = 6t^2 - 2t^3 + C = x(t)\)

Find the value of \(\displaystyle C\) by plugging \(\displaystyle t = 0\) into \(\displaystyle x(t)\).

\(\displaystyle 6\cdot0^2 - 2\cdot0^3 + C = 0\)

\(\displaystyle C = 0\)

Therefore

\(\displaystyle x(t) = 6t^2 - 2t^3\)

Evaluate this at \(\displaystyle t=2\) to find the position.

\(\displaystyle x(2) = 6\cdot2^2 - 2\cdot2^3\)

\(\displaystyle x(2) = 24 - 16\)

\(\displaystyle x(2) = 8\)

In order to find the position function from the velocity function we must take the integral of the velocity function since

\(\displaystyle s(t)=\int v(t)\,dt\).

When taking the integral, we use the trigonometric property which states

\(\displaystyle \int sec^2(x)\,dx = tan(x)\).

As such,

\(\displaystyle \int v(t)\, dt=\int sec^2 (t)\,dt\)

\(\displaystyle =tan(t)+c\)

Therefore the position function is

\(\displaystyle s(t)=tan(t)+c\).

If the function b(s) models the acceleration of a ping pong ball being thrown across a table, find the function which models the ball's position if the velocity when \(\displaystyle s=3\) is 108 and the position when \(\displaystyle s=5\) is 45.

\(\displaystyle b(s)=-15s^2+18s-8\)

We are asked to relate acceleration and position. This may seem confusing, but if we remember that acceleration is the first derivative of velocity, we know that we should start by integrating b(s).

\(\displaystyle \int b(s)=\int -15s^2+18s-8ds=-5s^3+9s^2-8s+c\)

Now we have the following:

\(\displaystyle B(s)=-5s^3+9s^2-8s+c\)

Which is almost our velocity function, but we need to use our given conditions to find c.

We are told that when s=3, the velocity is equal to 108. Let's plug those numbers in and solve for c!

\(\displaystyle 108=-5(3)^3+9(3)^2-8(3)+c\)

\(\displaystyle 108=-78+c\)

\(\displaystyle c=186\)

Which makes our velocity function:

\(\displaystyle B(s)=-5s^3+9s^2-8s+186\)

Now, from our velocity function we can integrate to get out position function. We know this because velocity is the first derivative of position. So...

\(\displaystyle \int B(s)ds=\int -5s^3+9s^2-8s+186ds\)

\(\displaystyle \int -5s^3+9s^2-8s+186ds=\frac{-5s^4}{4}+3s^3-4s^2+186s+c\)

We'll call our position function P(s), for position. 

\(\displaystyle P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s+c\)

We are almost there, but we know that \(\displaystyle P(5)=45\), so we need to find "c"

\(\displaystyle 45=\frac{-5(5)^4}{4}+3(5)^3-4(5)^2+186(5)+c\)

Simplify:

\(\displaystyle 45=423.75+c\)

\(\displaystyle c=-378.75\)

Making our position function:

\(\displaystyle P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s-378.75\)

In order to determine the position function from the velocity function, we need to integrate the velocity function.

\(\displaystyle s(t) = \int v(t)\: dt\)

Integrate the velocity function.

\(\displaystyle s(t)=\int v(t) \: dt= \int 2t\:dt = t^2 + C\)

Since the initial condition is zero, substitution of \(\displaystyle s(t)=0\) and \(\displaystyle t=0\) will leave \(\displaystyle C=0\), which will eliminate the constant term.  Rewrite the position function.

\(\displaystyle s(t)=t^2\)

Solve for \(\displaystyle s(t=2)\).

\(\displaystyle s(t=2) = 2^2 =4\)

To solve, integrate to find position and then plug in \(\displaystyle t\). Thus,

\(\displaystyle p(t)=3t^2+t=3(2)^2+2=14\)

To find the maximum height of the ball we will first need to find the moment which it has stoped moving. This is the same as finding when the velocity is equal to zero. The velocity at the maximum height will be \(\displaystyle 0\), so we need to determine when that happens. 

For this particular function we will need to apply the power rule which states, \(\displaystyle (x^n)'=nx^{n-1}\).

Applying the power rule to our position function we are able to find the velocity function.

\(\displaystyle \\v(t)=h'(t)=(-16t^2+160t+25)'\\ \\v(t)=(2*-16)t^{2-1}+(1*160)t^{1-1}+0\\ \\ v(t)=-32t^1+160t^0\\ \\v(t)=-32t+160(1)\\ \\v(t)=-32t+160\)

From here, set the velocity function equal to zero and solve for time.

\(\displaystyle 0=-32t+160\)

\(\displaystyle -160=-32t\)

\(\displaystyle 5=t\)

Now we can find the height of the ball at time \(\displaystyle t=5\).

\(\displaystyle h(5)=-16(5)^2+160(5)+25=-400+800+25=425\)

The original problem tells us that the calculations are in inches, so we need to convert to feet.

\(\displaystyle \frac{425}{12}\approx35.417\ feet\).

We are given very little information, so let us determine what we know. 

\(\displaystyle a=-32\ \frac{ft}{sec^2}\)

\(\displaystyle h(0)=300\ ft\)

\(\displaystyle v(0)=-5\ \frac{ft}{sec}\)

Note that both of these values are negative because the object is traveling downward. (This is personal preference. You can use down as the positive direction in this problem as well so long as the signs of the velocity and acceleration are the same.)

We know that we can integrate acceleration to derive velocity, so we can find \(\displaystyle v(t)\).

Recall the rules for integration that apply to this specific problem,

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\)

applying this rule to the accelaration we are able to find the velocity function.

\(\displaystyle v(t)=\int a(t)dt=\int -32dt=-32t+c\)

We can now use the initial value for \(\displaystyle v\) to determine \(\displaystyle c\).

\(\displaystyle v(t)=-32t+c\).

Plugging in \(\displaystyle v(0)=-5\ \frac{ft}{sec}\), we find

\(\displaystyle -5=-32(0)+c\)

\(\displaystyle -5=c\)

\(\displaystyle v(t)=-32t-5\).

Again, we can integrate the velocity function to find the position function.

\(\displaystyle \\h(t)=\int v(t)dt=\int (-32t-5)dt\\ h(t)=\frac{-32}{2}t^{1+1}-5t+d\)

\(\displaystyle h(t)=-16t^2-5t+d\).

We know that \(\displaystyle h(0)=300\), so we can use that to find \(\displaystyle d\).

\(\displaystyle 300=-16(0)^2-5(0)+d\)

\(\displaystyle 300=d\)

\(\displaystyle h(t)=-16t^2-5t+300\)

Now we simply need to determine \(\displaystyle t\) when the height of the bucket is \(\displaystyle 0\).

\(\displaystyle 0=-16t^2-5t+300\)

Using the quadratic formula,

\(\displaystyle t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

where \(\displaystyle a=-16, b=-5, c=300\)

\(\displaystyle t=\frac{--5\pm \sqrt{(-5)^2-4(-16)(300)}}{2(-16)}\)

we find that \(\displaystyle t\approx -4.489, 4.177\)

Because negative time does not make sense, our solution is \(\displaystyle 4.177\ seconds\).

Let us first determine what we are given.

\(\displaystyle x(t)=-\frac{1}{2}gt^2+v_0t_+s_0\)

\(\displaystyle x_0=0\ feet\)

\(\displaystyle v_0=128\ \frac{ft}{sec}\)

\(\displaystyle a(t)=32\ \frac{ft}{sec^2}\), \(\displaystyle (g=a)\)

Let's plug what we know into the general formula.

\(\displaystyle x(t)=-\frac{1}{2}(32)t^2+(128)t+0\)

\(\displaystyle x(t)=-16t^2+128t\)

The derivative of position will give us the velocity function. For this particular function we will need to use the power rule to find the derivative.

The power rule states, \(\displaystyle (x^n)'=nx^{n-1}\).

Applying the power rule to the position function we are able to find the following velocity function.

\(\displaystyle \\v(t)=x'(t)=(-16t^2+128t)\\ \\v(t)=(2*-16)t^{2-1}+(1*128)t^{1-1}\\ \\ v(t)=-32t+128t^0=-32t+128(1)\\ \\v(t)=-32t+128\)

When the arrow reaches its maximum height, the velocity will be \(\displaystyle 0\). Therefore, set the velocity function that was found equal to zero and solve for time.

\(\displaystyle 0=-32t+128\)

\(\displaystyle -128=-32t\)

\(\displaystyle t=4\) seconds.

Now we can find the position of the arrow at \(\displaystyle t=4\) seconds.

\(\displaystyle x(4)=-16(4)^2+128(4)=256\ feet\).

To solve, simply integrate the following function and plug in \(\displaystyle t=1\).

\(\displaystyle p(t)=t^3+t^2+t\)

\(\displaystyle p(1)=1^3+1^2+1=3\)