Velocity is the derivative of position. The derivative of \(\displaystyle x^{n}\) is \(\displaystyle nx^{n-1}\).

With this information the velocity function can be found.

\(\displaystyle v(t)=60t^3+15t^2-1\) 

In order to solve this problem, you must know that the derivative of a position function is the veloctiy function and the derivative fo the velocity function is the acceleration function. Thefore, by taking the double integral of the acceleration function we will be able to find the position function.  

In order to take the integral of the acceleration function, \(\displaystyle a(t)=4.9\), we must first know the general rule of integration \(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\).   

Taking the integral of the acceleration function, we find that the veloctiy function is equivalent to \(\displaystyle v(t)=4.9t+C\).  We must somehow figure out how to solve for \(\displaystyle C\).  

Taking the integral of the veloctiy function to get the position function, we find that the position function is \(\displaystyle S(t)=2.45t^2+Ct\).  Note that this time another \(\displaystyle C\) isn't included because the intial position of the obejct is 0.

We know that the object traveled 500 meters in 10 seconds, therefore by plugging in \(\displaystyle S(10)=500\)  we can solve for \(\displaystyle C\).

\(\displaystyle 500=2.45(10)^2+10C\)

\(\displaystyle C=25.5\)

Therefore the veloctiy function becomes \(\displaystyle v(t)=4.9t+25.5\).  The problem asks us to find the veloctiy of the object at 5 seconds, plugging in 5 we find that the velocity of the object is \(\displaystyle 50\frac{m}{s}\).

Velocity is equal to the derivative of position with respect to time.

The derivative of this position function, by the power rule which states, 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}\) is 

\(\displaystyle \\s(t)=t^2 -6t +5 \\s'(t)=2t^{2-1}-6t^{1-1}\)

\(\displaystyle v(t)=s'(t) = 2t - 6\).

Average velocity is defined as \(\displaystyle v_{avg} = \frac{\Delta s}{\Delta t}\).

Therefore, on the interval \(\displaystyle 0 \leq t \leq 5\), 

\(\displaystyle \\v_{avg} = \frac{s(5)-s(0)}{5-0}\\ v_{avg} = \frac{20 - 15}{5}\\ v_{avg} = 1\)

The instantaneous velocity can be found by taking the derivative of the position function.

Recall the following rules of differentiation to help solve this problem.

Power Rule:

 \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}\) 

Natural Log Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(aln(bt)) = \frac{a}{t}\)

The instantaneous velocity at \(\displaystyle t=4\), by the power rule and the rule for natural logs, is 

\(\displaystyle \\v(4) = s'(4) \\ v(t) = s'(t) = \frac{3}{t}+\frac{3t^2}{16}\\ v(4) = \frac{3}{4} + 3\\ \\v(4) = 3.75\)

Velocity is the derivative of position with respect to time.

Recall the following rule of differentiation to help solve this problem.

Power Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}\) 

Therefore, by the power rule, remembering that \(\displaystyle \frac{3}{t} = 3t^{-1}\) and \(\displaystyle \sqrt{t} = t^{1/2}\), the velocity function is 

\(\displaystyle s'(t) = v(t) = -\frac{3}{t^2} + \frac{1}{2\sqrt{t}}\).

Velocity is equal to the derivative of position with respect to time.

Recall the following rules of differentiation to help solve this problem.

Power Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}\) 

Differentiation rules for sine and cosine: 

\(\displaystyle \\\frac{\mathrm{d} }{\mathrm{d} t}(asin(bt)) = abcos(bt)\\ \frac{\mathrm{d} }{\mathrm{d} t}(acos(bt)) = -absin(bt)\\\)

Product rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(u(t)v(t)) = u(t)v'(t)+u'(t)v(t)\)

Since \(\displaystyle s(t) = tcos(t)\), then, by the product rule, the power rule, and the differentiation rules for sine and cosine, 

\(\displaystyle v(t) = cos(t) - tsin(t)\).

The velocity is just the derivative of the position.

The power rule states that the derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\).

This is the only rule we need to find the derivative of the position function.

\(\displaystyle v(t)=12x^2-10x-1\)

Here we need the average velocity.

An average velocity can be found by taking the integral of the velocity function then taking the difference of that function at the given bounds and dividing that by the interval. In other words:

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

We are given the velocity function, so putting the pieces together we get:

\(\displaystyle V_{avg}=\frac{1}{\frac{\pi}{4}--\frac{\pi}{4}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5+tan(t)\)

\(\displaystyle V_{avg}=\frac{1}{\frac{\pi}{2}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5+tan(t)\)

 \(\displaystyle V_{avg}={\frac{2}{\pi}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5+tan(t)\)

We will use the power rule to integrate the constant which states, 

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

To integrate tangent we will use the rules of trigonometric functions which states that,

\(\displaystyle \int tan(x)=-ln|cos(x)|\).

Applying these rules we get the following:

\(\displaystyle V_{avg}=\frac{2}{\pi}\left[5t+-ln|cos(t)|\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)

Now plugging in our initial and final values we get our answer, 

\(\displaystyle V_{avg}=\frac{2}{\pi}\left[\left[5\left(\frac{\pi}{4} \right )+-ln\left|cos\left(\frac{\pi}{4} \right )\right|\right]-\left[5\left(\frac{-\pi}{4} \right )+-ln\left|cos\left(\frac{-\pi}{4} \right )\right|\right]\right]\)

\(\displaystyle V_{avg}=\frac{2}{\pi}\left[\left[\left(\frac{5\pi}{4} \right )+-ln\left|\frac{1}{\sqrt2} \right|\right]-\left[\left(\frac{-5\pi}{4}\right)+-ln\left|\frac{1}{\sqrt2}\right|\right]\right]\)

\(\displaystyle V_{avg}=\frac{2}{\pi}\left[\frac{10\pi}{4} \right ]=\frac{20}{4}=5\)

Velocity is the derivative of position. The derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\).

That means that we can get the velocity fucntion with this information.

\(\displaystyle v(t)=3t^2-10t+10\)

To get the velocity at t=2 we must plug in 2 for t in the velocity function.

\(\displaystyle v(2)=3(2)^2-10(2)+10\)

\(\displaystyle =2\)