Velocity is the time derivative of position:

\(\displaystyle v(t)=\frac{d}{dt}p(t)\)

In the case of two position functions for the description of two-dimensional location, there will simply be two corresponding velocity functions:

For the position functions

\(\displaystyle x(t)=t^2+3t-4\)

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

The velocity functions are then:

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

\(\displaystyle v_y(t)=6x(t)\frac{dx(t)}{dt}\)

\(\displaystyle v_y(t)=6x(t)v_x(t)\)

\(\displaystyle v_y(t)=6(t^2+3t-4)(2t+3)\)

At time \(\displaystyle t=3\)

\(\displaystyle v_x(t)=6+3=9\)

\(\displaystyle v_y(t)=6(9+9-4)(6+3)=6(14)(9)=756\)

Velocity can be found as the time derivative of position:

\(\displaystyle v(t)=\frac{d}{dt}p(t)\)

In the case of three-dimensional descriptions, that simply means there are three velocity functions. For the positional functions

\(\displaystyle x(t)=3cos(\frac{1}{2}t)\)

\(\displaystyle y(t)=2sin(t)cos(t)\)

\(\displaystyle z(t)=t^2+t\)

The resultant velocity functions are:

\(\displaystyle v_x(t)=-\frac{3}{2}sin(\frac{1}{2}t)\)

\(\displaystyle v_y(t)=2cos^2(t)-2sin^2(t)\)

\(\displaystyle v_z(t)=2t+1\)

At time \(\displaystyle t=\pi\)

\(\displaystyle v_x(\pi)=-\frac{3}{2}sin(\frac{1}{2}\pi)=-\frac{3}{2}\)

\(\displaystyle v_y(\pi)=2cos^2(\pi)-2sin^2(\pi)=2\)

\(\displaystyle v_z(\pi)=2\pi+1\)

The magnitude of the velocity is then the root of the sum of the squares:

\(\displaystyle v(\pi)=\sqrt{(-\frac{3}{2})^2+(2)^2+(2\pi+1)^2}\)

\(\displaystyle v(\pi)=\sqrt{\frac{9}{4}+4+4\pi^2+4\pi+1}\)

\(\displaystyle v(\pi)=\sqrt{4\pi^2+4\pi+\frac{29}{4}}\)

Velocity can be found by integrating acceleration with respect to time:

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

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

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

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

For the acceleration function 

\(\displaystyle a(t)=9t^2+15e^{-3t}\),

the velocity function is thus:

\(\displaystyle v(t)=3t^3-5e^{-3t}+C\)

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

\(\displaystyle v(0)=3(0^3)-5e^{-3(0)}+C=10\)

\(\displaystyle C=15\)

From here, the full velocity function can be found:

\(\displaystyle v(t)=3t^3-5e^{-3t}+15\)

\(\displaystyle v(5)=3(5^3)-5e^{-3(5)}+15\)

\(\displaystyle v(5)=390-5e^{-15}\)

Velocity can be found as the time derivative of position:

\(\displaystyle v(t)=\frac{d}{dt}p(t)\)

For the three position functions

\(\displaystyle x(t)=2t+4\)

\(\displaystyle y(t)=x^3(t)\)

\(\displaystyle z(t)=2y^2(t)\)

There can be found three velocity funcitons:

\(\displaystyle v_x(t)=2\)

\(\displaystyle v_y(t)=3x^2(t)\frac{dx(t)}{dt}\)

\(\displaystyle v_z(t)=4y(t)\frac{dy(t)}{dt}\)

Now note that these derivative terms correspond to velocities, i.e. \(\displaystyle v_x(t)=\frac{dx(t)}{dt}\) and \(\displaystyle v_y(t)=\frac{dy(t)}{dt}\).

Knowing this, we can replace terms in each function so that each velocity is in terms of time:

\(\displaystyle v_x(t)=2\)

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

\(\displaystyle v_z(t)=4(2t+4)^3[3(2t+4)^2(2)]\)

Therefore at time \(\displaystyle t=2\)

\(\displaystyle v_x(2)=2\)

\(\displaystyle v_y(2)=3(4+4)^2(2)=384\)

\(\displaystyle v_z(2)=4(4+4)^3(384)=786432\)

Velocity can be found as the time derivative of position:

\(\displaystyle v(t)=\frac{d}{dt}p(t)\)

For the two position functions

\(\displaystyle x(t)=4cos(\frac{1}{2}t)+sin(3t)\)

\(\displaystyle y(t)=2sin(\frac{1}{2}t)cos(\frac{3}{2}t)\)

The velocity functions are found to be:

\(\displaystyle v_x(t)=-2sin(\frac{1}{2}t)+3cos(3t)\)

\(\displaystyle v_y(t)=cos(\frac{1}{2}t)cos(\frac{3}{2}t)-3sin(\frac{1}{2}t)sin(\frac{3}{2}t)\)

The velocities at time \(\displaystyle t=\pi\) are found to be:

\(\displaystyle v_x(\pi)=-2sin(\frac{1}{2}\pi)+3cos(3\pi)=-2-3=-5\)

\(\displaystyle v_y(\pi)=cos(\frac{1}{2}\pi)cos(\frac{3}{2}\pi)-3sin(\frac{1}{2}\pi)sin(\frac{3}{2}\pi)\)

\(\displaystyle v_y(\pi)=0(0)-3(1)(-1)=3\)

The magnitude of the velocity can be found by taking the root of the sum of squares:

\(\displaystyle v(\pi)=\sqrt{(-5)^2+(3)^2}\)

\(\displaystyle v(\pi)=\sqrt{34}\)

To find the velocity function \(\displaystyle v(t)\), we need to integrate the acceleration function \(\displaystyle a(t)\).

To integrate this function, we need the following formulae:

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

\(\displaystyle v(t)=\int a(t) = \int 3t^{2}+2t = \frac{3t^{3}}{3}+\frac{2t^2}{2}+C=t^{3}+t^{2}+C\)

Now, to solve for the constant, we use the initial conditions:

Since \(\displaystyle v(0)=3\), we can plug in \(\displaystyle 0\) for \(\displaystyle t\) in the velocity function and set it equal to \(\displaystyle 3\):

\(\displaystyle v(0)=3(0)^{2}+2(0)+C=0+0+C=3\)

Therefore, \(\displaystyle C=3\)  

So, the specific solution for the velocity is \(\displaystyle v(t)=t^{3}+t^{2}+3\).

Average velocity can be found as the total distance travelled over the total time expended.

Distance travelled can be found by integrating the velocity function over the specified interval of time \(\displaystyle t=[a,b]\).

Therefore

\(\displaystyle v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt\)

For the velocity function

\(\displaystyle v(t)=3t^2+4t+6\)

\(\displaystyle v_{avg}=\frac{1}{10-2}\int_2^{10} (3t^2+4t+6)dt\)

\(\displaystyle v_{avg}=\frac{1}{8}(t^3+2t^2+6t)|_2^{10}\)

\(\displaystyle v_{avg}=\frac{1}{8}[(1000+200+60)-(8+8+12)]\)

\(\displaystyle v_{avg}=\frac{1}{8}[(1260)-(28)]\)

\(\displaystyle v_{avg}=\frac{1}{8}(1232)\)

\(\displaystyle v_{avg}=154\)

Average velocity can be found by dividing total distance traveled over total time elapsed.

When given a velocity function, distance traveled over an interval of time \(\displaystyle t=[a,b]\) can be found by integrating the velocity over this interval. By dividing by the length of this interval, an average velocity can be found:

\(\displaystyle v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt\)

For the velocity function

\(\displaystyle v(t)=t^2+t+1\)

The average velocity over the interval \(\displaystyle t=[0,3]\) can be found as

\(\displaystyle v_{avg}=\frac{1}{3-0}\int_0^3(t^2+t+1)dt\)

\(\displaystyle v_{avg}=\frac{1}{3}(\frac{t^3}{3}+\frac{t^2}{2}+t)|_0^3\)

\(\displaystyle v_{avg}=\frac{1}{3}[(9+\frac{9}{2}+3)-(0+0+0)]\)

\(\displaystyle v_{avg}=\frac{11}{2}=5.5\)

The particle changes directions when its velocity undergoes a sign change. Velocity can be found as the time derivative of position, so for the position function

\(\displaystyle x(t)=t^2-10t+24\)

The velocity can be found as

\(\displaystyle v(t)=2t-10\)

The particle initially travels in the negative direction, since \(\displaystyle v(0)=-10\). The velocity will change in sign after the point that the function reaches zero, if such a point exists for a positive time value:

\(\displaystyle 2t-10=0\)

\(\displaystyle 2t=10\)

\(\displaystyle t=5\)

This indicates the turning point. All later times, the particle will move in a positive direction.

Velocity can be found by taking the derivative of position with respect to time. For the position function

\(\displaystyle x(t)=-t^3+9t^2+5t+20\)

The velocity function is

\(\displaystyle v(t)=-3t^2+18t+5\)

Now to find where the velocity has minima or maxima, take the derivative again with respect to time to find acceleration, and find when the accerlation is zero:

\(\displaystyle a(t)=-6t+18\)

\(\displaystyle -6t+18=0\)

\(\displaystyle t=3\)

Prior to this time, the acceleration is postive, and after this time the acceleration is negative, so this indicates a maximum.

The maximum velocity is then

\(\displaystyle v_{max}=v(3)=-27+54+5\)

\(\displaystyle v_max=32\)