To find the velocity function, we need to find the derivative of the position function.

So lets take the derivative of \(\displaystyle p(t)=t^2+t\) with respect to \(\displaystyle t\).

The derivative of \(\displaystyle t^2\) is \(\displaystyle 2t\) because of Power Rule:

\(\displaystyle f(x)=a^n \rightarrow \ f'(x)=na^{n-1}\)

\(\displaystyle p(t)=t^2 \rightarrow p'(t)=2t^{2-1}=2t\)

The derivative of \(\displaystyle t\) is \(\displaystyle 1\) due to Power Rule

\(\displaystyle p(t)=t \rightarrow p'(t)=1t^{1-1}=1t^0=1(1)=1\)

So...

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

To find the velocity at \(\displaystyle t=2\) we just plug \(\displaystyle t\) into the veloctiy function

\(\displaystyle v(2)=4\cdot 2+3\)

\(\displaystyle v(2)=8+3\)

\(\displaystyle v(2)=11\)

To find the veloctiy function, we need to take the derivative of the position function

So \(\displaystyle p(t)=3t^2+4\) would turn into \(\displaystyle v(t)=6t\)

because of the Power Rule as well as the rule that the derivative of a constant is \(\displaystyle 0\).

\(\displaystyle p(t)=a^n \rightarrow v(t)=na^{n-1}\)

\(\displaystyle v(t)=2\cdot 3t^1+0=6t\)

To find the velocity function, we need to take the derivative of the position function in relation to \(\displaystyle t\)

So

\(\displaystyle p(t)=t^3+4t\)

changes into

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

using the Power Rule

\(\displaystyle p(t)=a^n \rightarrow v(t)=na^{n-1}\)

\(\displaystyle v(t)=3 \cdot 1t^{3-1}+4t^{1-1}=3t^2+4t^0=3t^2+4\).

To find the velocity at time \(\displaystyle t=6\) we plug in for \(\displaystyle t\)

So

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

becomes

\(\displaystyle v(6)=6^2+3\cdot 6\)

\(\displaystyle v(6)=36+18\)

\(\displaystyle v(6)=54\).

To find the velocity function we need to take the derivative of the position function

So \(\displaystyle p(t)=t^4+t^2+2t\) turns into \(\displaystyle v(t)=4t^3+2t+2\) because of power rule.

\(\displaystyle p(t)=a^n \rightarrow v(t)=na^{n-1}\)

\(\displaystyle v(t)=4t^{4-1}+2t^{2-1}+2t^{1-1}=4t^3+2t+2\)

We are given a function dealing with distance and asked to find a velocity. recall that velocity is the first derivative of position. Find the first derivative of h(t) and evaluate at t=15.

\(\displaystyle h(t)=\frac{4t^3}{3}+\frac{4.5t^2}{3}+3t-7\)

\(\displaystyle h'(t)=4t^2+3t+3\)

\(\displaystyle h'(15)=4*15^2+3*15+3=900+45+3=948\)

\(\displaystyle 948\frac{m}{s}\)

Recall that velocity is the first derivative of position.

To find the velocity after 3 seconds, we need p'(3).

\(\displaystyle p(t)=6t^4-5t^3+6t-18\)

\(\displaystyle p'(t)=24t^3-15t^2+6\)

\(\displaystyle p'(3)=24\cdot3^3-15\cdot3^2+6=519\)

To find the velocity for the long run, we need to find the limit as t gets bigger and bigger.

Note that to find this limit, we need to take the limit componentwise.

We know the following:

\(\displaystyle \lim_{t\rightarrow \infty}\frac{t^2}{t^2+1}=1\)

\(\displaystyle \lim_{t\rightarrow \infty}e^{-t^2}=0\)

\(\displaystyle \lim_{t\rightarrow \infty}artan(t^2)=\frac{\pi}{2}\).

Therefore as t becomes bigger we will reach the velocity given by:

\(\displaystyle \left(1,0,\frac{\pi}{2}\right)\)

To find who is driving faster, we need to compare the norm of their velocities:

We will call \(\displaystyle \vec{V_{1}},\vec{V_{2}}\)  the velocities of the first and the second drivers respectively.

Using the Chain Rule on the position function we have the following velocities,

\(\displaystyle \vec{V_{1}}=(-sin(t),cos(t))\).

\(\displaystyle \vec{V_{2}}=(-2sin(2t),2cos(2))\).

Now we find the speed of both drivers.

\(\displaystyle v_{1}=\sqrt{sin^2(t)+cos^2(t)}=1\)

Use the trigonometric identity to find velocity two.

\(\displaystyle v_{2}=\sqrt{4sin^2(t)+4cos^2(t)}=\sqrt{4\left( \frac{1}{2}-\frac{1}{2}cos(2x)\right)+4\left(\frac{1}{2}+\frac{1}{2}cos(2x)\right)}\)

\(\displaystyle v_2=\sqrt{2-2cos(2x)+2+2cos(2x)}=\sqrt{2+2}=\sqrt{4}= 2\)

and we see that the second driver is driving faster.