The position function is equal to the integral of the velocity function:

\(\displaystyle \int (23t^2+e^t+20)dt=\frac{23t^3}{3}+e^t+20t+C\)

and was found using the following rules:

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\), \(\displaystyle \int e^xdx=e^x+C\)

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

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

When taking the integral, we will use the inverse power rule which states,

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

Applying this rule to each term we get

\(\displaystyle \int v(t)\,dt = \int 4t^2+15\,dt\)

\(\displaystyle = \frac{4t^3}{3}+15t+c\). 

As such,

\(\displaystyle s(t)=\frac{4}{3}t^3+15t+c\).

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

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

When taking the integral, we will use the trigonometric integral,

 \(\displaystyle \int cos\,x \, dx = sin\,x+c\).

Applying this rule we get

\(\displaystyle \int v(t)\,dt = \int cos\,t\,dt\) 

 \(\displaystyle =sin\,t+c\).

As such,

\(\displaystyle s(t)= sin\, t + c\).

Position can be found by integrating acceleration with respect to time twice (or integrating velocity with respect to time once):

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

For the acceleration function

\(\displaystyle a(t)=6t+24\)

First find velocity:

Use the following rule to find velocity,

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

\(\displaystyle v(t)=3t^2+24t+C_1\)

To find the constant of integration, use the initial velocity condition:

\(\displaystyle v(0)=5=3(0^2)+24(0)+C_1\)

\(\displaystyle C_1=5\)

\(\displaystyle v(t)=3t^2+24t+5\)

Now integrate once more to find position:

\(\displaystyle x(t)=t^3+12t^2+5t+C_2\)

To find this second contant, use the inital position:

\(\displaystyle x(0)=10=0^3+12(0^2)+5(0)+C_2\)

\(\displaystyle C_2=10\)

\(\displaystyle x(t)=t^3+12t^2+5t+10\)

\(\displaystyle x(3)=27+108+15+10\)

\(\displaystyle x(3)=160\)

The position function can be found by integrating the velocity function with respect to time:

\(\displaystyle x(t)=\int v(t)dt\)

For the velocity function \(\displaystyle v(t)=cos(\frac{1}{2}t)-sin(t)+2\), the position function is therefore:

\(\displaystyle x(t)=2sin(\frac{1}{2}t)+cos(t)+2t+C\)

To find the constant of integration, use the initial position:

\(\displaystyle x(0)=2sin(\frac{1}{2}0)+cos(0)+2(0)+C=0\)

\(\displaystyle 1+C=0\)

\(\displaystyle C=-1\)

\(\displaystyle x(t)=2sin(\frac{1}{2}t)+cos(t)+2t-1\)

\(\displaystyle x(\pi)=2sin(\frac{1}{2}\pi)+cos(\pi)+2(\pi)-1\)

\(\displaystyle x(\pi)=2-1+2(\pi)-1\)

\(\displaystyle x(\pi)=2\pi\)

The position function can be found by integrating the velocity function with respect to time:

\(\displaystyle x(t)=\int v(t)=\int (\frac{1}{2}te^{t^2}-sin(\frac{1}{3}t))dt\)

To find the integral we will need to apply a few different rules.

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

\(\displaystyle \int e^u=e^u\cdot \frac{1}{\frac{du}{dx}}\)

Thus we get the following position function.

\(\displaystyle x(t)=\frac{1}{4}e^{t^2}+3cos\left(\frac{1}{3}t\right)+C\)

The constant of integration can be found by using the initial condition:

\(\displaystyle x(0)=\frac{1}{4}e^{0^2}+3cos\left(\frac{1}{3}(0)\right)+C=0\)

\(\displaystyle \frac{1}{4}+3+C=0\)

\(\displaystyle C=-\frac{13}{4}\)

Knowing this provides the complete position function:

\(\displaystyle x(t)=\frac{1}{4}e^{t^2}+3cos\left(\frac{1}{3}t\right)-\frac{13}{4}\)

\(\displaystyle x(2)=\frac{1}{4}e^4+3cos\left(\frac{2}{3}\right)-\frac{13}{4}\)

Position can be found by integrating velocity with respect to time:

\(\displaystyle p(t)=\int v(t)dt\)

For velocities:

\(\displaystyle v_x(t)=t^2+sin(t)\)

\(\displaystyle v_y(t)=tcos(t^2)+3\)

The position functions are:

\(\displaystyle x(t)=\frac{t^3}{3}-cos(t)+C_x\)

\(\displaystyle y(t)=\frac{sin(t^2)}{2}+3t+C_y\)

These constants of integration can be found by using the given initial conditions:

\(\displaystyle x(0)=\frac{0^3}{3}-cos(0)+C_x=0\)

\(\displaystyle C_x=1\)

\(\displaystyle y(0)=\frac{sin(0^2)}{2}+3(0)+C_y=0\)

\(\displaystyle C_y=0\)

Which gives the definite integrals:

\(\displaystyle x(t)=\frac{t^3}{3}-cos(t)+1\)

\(\displaystyle y(t)=\frac{sin(t^2)}{2}+3t\)

\(\displaystyle x(2\pi)=\frac{(2\pi)^3}{3}-cos(2\pi)+1\)

\(\displaystyle x(2\pi)=\frac{8\pi^3}{3}\)

\(\displaystyle y(2\pi)=\frac{sin((2\pi)^2)}{2}+3(2\pi)\)

\(\displaystyle y(2\pi)=6\pi\)

Position can be found by integrating with respect to time velocity once, or integrating with respect to time acceleration twice:

\(\displaystyle p(t)=\int v(t)dt= \int \int a(t)dt\)

Begin by finding velocity functions. For the acceleration functions

\(\displaystyle a_x(t)=6t+4\)

\(\displaystyle a_y(t)=12t^2+2\)

Velocities are:

\(\displaystyle v_x(t)=3t^2+4t+C_{1x}\)

\(\displaystyle v_y(t)=4t^3+2t+C_{1y}\)

Find the constants of integration by using the given initial conditions:

\(\displaystyle v_x(0)=3(0^2)+4(0)+C_{1x}=2\)

\(\displaystyle C_{1x}=2\)

\(\displaystyle v_y(0)=4(0^3)+2(0)+C_{1y}=1\)

\(\displaystyle C_{1y}=1\)

Plugging these values back in gives the definite velocity functions:

\(\displaystyle v_x(t)=3t^2+4t+2\)

\(\displaystyle v_y(t)=4t^3+2t+1\)

Now, find the position functions by integrating once more:

\(\displaystyle x(t)=t^3+2t^2+2t+C_{2x}\)

\(\displaystyle y(t)=t^4+t^2+t+C_{2y}\)

Find these new constants of integration the same way as before:

\(\displaystyle x(0)=0^3+2(0^2)+2(0)+C_{2x}=3\)

\(\displaystyle C_{2x}=3\)

\(\displaystyle y(0)=0^4+0^2+0+C_{2y}=4\)

\(\displaystyle C_{2y}=4\)

And the definite position functions are thus:

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

\(\displaystyle y(t)=t^4+t^2+t+4\)

Then at time \(\displaystyle t=3\), the particle's position is:

\(\displaystyle x(3)=3^3+2(3)^2+2(3)+3=27+18+6+3=54\)

\(\displaystyle y(3)=3^4+3^2+3+4=81+9+3+4=97\)

To find the position function for the race car, we must integrate the velocity function.

To integrate this function, we must use the following formula:

\(\displaystyle \int cf(x)dx=c\int f(x)dx, \int x^{n}dx=\frac{1}{n+1}x^{n+1}\)

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

Now, to find the position at time \(\displaystyle t=4\), we plug in \(\displaystyle t=4\) into the position function.

\(\displaystyle p(4)=\frac{256}{4}-\frac{128}{3}+16 = 37\frac{1}{3} meters\)

To find when it will get back to Tom. Assume Tom is at position 0, and find where the position function equals zero. Position funtion is the integral of the velocity funtion which is the intergral of the acceleration function. 

\(\displaystyle a(t)=-4\)

\(\displaystyle v(t)=\int-4dt=-4x+v(0)\)

using, 

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

\(\displaystyle v(0)=10\) by what's given.

\(\displaystyle v(t)=-4t+10\)

\(\displaystyle p(t)=\int(-4t+10)dt=-2t^2+10t+p(0)\)

\(\displaystyle p(0)=0\) since we assume Tom is at position 0, so now we must find where it is 0 again.

\(\displaystyle -2t^2+10t=0 \Rightarrow -2t(t+5)=0 \Rightarrow t=0,5\)

Thus, it will return to Tom at posistion after 5 seconds.