Velocity is the derivative of position, so in order to obtain an equation for position, we must integrate the given equation for velocity:

\(\displaystyle s(t)=\int v(t)dt=\int (7t+4)dt=\frac{7}{2}t^2+4t+C\)

The next step is to solve for C by applying the given initial condition, s(0)=5:

\(\displaystyle s(t)=\frac{7}{2}t^2+4t+C\)

\(\displaystyle s(0)=0+0+C=5\rightarrow C=5\)

So our final equation for position is:

\(\displaystyle s(t)=\frac{7}{2}t^2+4t+5\)

To find the position of the ball, we plug in \(\displaystyle t=1\)

So \(\displaystyle p(t)=4t+2\) turns into:

\(\displaystyle p(1)=4\cdot 1+2\)

\(\displaystyle p(1)=4+2\)

\(\displaystyle p(1)=6\)

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

We are starting with the following

\(\displaystyle \small \small v(t)=t^5+3t^4-\frac{7t^2}{3}\)

We need to perform the following:

\(\displaystyle \small \small \small \small\int v(t)dt=\int t^5+3t^4-\frac{7t^2}{3}dt\)

Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.

\(\displaystyle \small \small \int t^5+3t^4-\frac{7t^2}{3}dt=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c\)

Which makes our position function, h(t), the following:

\(\displaystyle \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c\)

Recall that velocity is the first derivative of position, so to find the position function we need to integrate \(\displaystyle v(t)\).

\(\displaystyle V(t)=\int 4t^3-5t^2+6dt\)

Becomes,

\(\displaystyle V(t)=t^4-\frac{5t^3}{3}+6t+c\)

Then, we need to find \(\displaystyle V(10)\) 

\(\displaystyle V(10)=10^4-\frac{5\cdot10^3}{3}+6\cdot10+c=8393.3\bar{3}+c\)

So our final answer is:

\(\displaystyle 8393.3\bar{3}+c \text{ }\frac{m}{s}\)

To find the position given the velocity curve, take the antiderivative of \(\displaystyle v(t)=30t^2\).

\(\displaystyle s(t)=\int v(t)dt=30\int t^2=30\left[\frac{t^{2+1}}{2+1} \right ]=30\left[\frac{t^3}{3} \right ]=\frac{30t^3}{3}=10t^3\)

Solve for,

 \(\displaystyle s(t=2)\).

\(\displaystyle s(t=2)=10(2^3)=80\)

To find the position from the velocity function, integrate 

\(\displaystyle v(t)=3t^5+4t\) by increasing the exponent of each t term and then dividing that term by the new exponent value.

\(\displaystyle s(t)=\int v(t)dt= \frac{3}{6}t^6+2t^2=\frac{1}{2}t^6+2t^2\)

To find the position function given the velocity, the velocity function needs to be integrated.

\(\displaystyle s(t)=\int v(t)dt= \int (6t+2)dt\)

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

Solve for \(\displaystyle s(t=2)\).

\(\displaystyle s(t=2)=3(2)^2+2(2)= 12+4=16\)

Integrate \(\displaystyle a(t)=2t-1\) to obtain the velocity function, \(\displaystyle v(t)\).

\(\displaystyle v(t)=\int a(t)dt=\int (2t-1) dt=t^2-t\)

Since velocity is 6, \(\displaystyle v(t)=6\).  Substitute and solve for time.

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

\(\displaystyle 0=t^2-t-6\)

\(\displaystyle 0=(t-3)(t+2)\)

\(\displaystyle t=3,-2\)

Since negative time does not exist, the only valid solution for t at this velocity is \(\displaystyle t=3\).

Integrate the velocity function \(\displaystyle v(t)\) to find the position function, \(\displaystyle s(t)\).

\(\displaystyle s(t)=\int v(t)dt=\frac{t^3}{3}-\frac{t^2}{2}\)

Substitute \(\displaystyle t=3\) to \(\displaystyle s(t)\) to obtain the position.

\(\displaystyle \frac{3^3}{3}-\frac{3^2}{2}= \frac{27}{3}-\frac{9}{2}=9-\frac{9}{2}=\frac{9}{2}\)

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

\(\displaystyle v(t)=5t^4-343t^6+51t^2\)

\(\displaystyle p(t)=V(t)=\int 5t^4-343t^6+51t^2dt\)

So our answer is:

\(\displaystyle p(t)=t^5-49t^7+17t^3+c\)

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

\(\displaystyle v(t)=5t^4-343t^6+51t^2\)

\(\displaystyle p(t)=V(t)=\int 5t^4-343t^6+51t^2dt\)

So we get:

\(\displaystyle p(t)=t^5-49t^7+17t^3+c\)

What we ultimately need is p(5), but first we need to find c: Use the point (2,2)

\(\displaystyle p(t)=t^5-49t^7+17t^3+c\)

\(\displaystyle 2=2^5-49\cdot 2^7+17\cdot 2^3+c\)

\(\displaystyle c=6106\)

So our position function is:

\(\displaystyle p(t)=t^5-49t^7+17t^3+6106\)

\(\displaystyle p(5)=5^5-49\cdot 5^7+17\cdot 5^3+6106=-3816769\)