Write the average velocity equation given the initial and final velocities.

\(\displaystyle v_f=\frac{y(t_f)-y(t_i)}{t_f-t_i}\)

Solve for \(\displaystyle y(t_f)\).

\(\displaystyle y(t_f=7)= 2(7)+1=15\)

Solve for \(\displaystyle y(t_i)\).

\(\displaystyle y(t_i=2)= 2(2)+1=5\)

Substitute the knowns into the average velocity equation.

\(\displaystyle v_f=\frac{y(t_f)-y(t_i)}{t_f-t_i}=\frac{15-5}{7-2}=\frac{10}{5}=2\)

Do not confuse average velocity with instantaneous velocity.  To determine the instantaneous velocity, take the derivative of the position function to obtain the velocity function.

\(\displaystyle s(t)= 5ln(t)+4t\)

\(\displaystyle v(t)= \frac{5}{t}+4\)

Substitute \(\displaystyle t=1\).

\(\displaystyle v(t)= \frac{5}{1}+4=9\)

Velocity can be viewed at the derivative of position, i.e. the rate of change of a position function.

So, we can find velocity by finding the derivative of the position function.

The position function is given as \(\displaystyle y = 2x\), so the derivative of the position function, using the power rule is,

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

 \(\displaystyle y=2x\rightarrow y'=1\cdot 2x^{1-1}=2x^0=2\), a constant, and the velocity function is thus always \(\displaystyle 2\).

By definition, velocity is the first derivative of a position function.

Therefore, \(\displaystyle v(t)=p'(t)\).

For this particular problem we will use the power rule to find the derivative. The power rule states,

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

Since 

\(\displaystyle p(t)=7t^{2}-2t+5\),  then applying the power rule we find 

\(\displaystyle v(t)=p'(t)=14t-2\).

Plugging in \(\displaystyle t=3\) gets us,

 \(\displaystyle v(t)=p'(t)=14(3)-2=42-2=40\).

By definition, velocity is the first derivative of a position function.

Therefore, \(\displaystyle v(t)=p'(t)\).

For this particular problem we will use the power rule to find the derivative. The power rule states,

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

Since \(\displaystyle p(t)=4t^{2}+9t+13\), then applying the power rule we find \(\displaystyle v(t)=p'(t)=8t+9\).

Plugging in \(\displaystyle t=4\) gets us \(\displaystyle v(t)=p'(t)=8(4)+9=32+9=41\).

By definition, velocity is the first derivative of a position function. Therefore, \(\displaystyle v(t)=p'(t)\).

For this particular problem we will use the power rule to find the derivative.

The power rule states,

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

Since \(\displaystyle p(t)=8t^{2}-6t+5\), then applying the power rule we find \(\displaystyle v(t)=p'(t)=16t-6\).

Plugging in \(\displaystyle t=7\) gets us \(\displaystyle v(t)=p'(t)=16(7)-6=112-6=106\).

Velocity is the first derivative of position, or \(\displaystyle v(t)=p'(t)\).

Given, 

\(\displaystyle p(t)=4t^{2}+9t-7\) we can use the power rule which states, 

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\).

Applying this rule we can deduce that, 

\(\displaystyle v(t)=p'(t)=8t+9\) .

Swapping in \(\displaystyle t=7\), we get 

\(\displaystyle v(7)=8(7)+9=56+9=65\).

Velocity is the first derivative of position, or \(\displaystyle v(t)=p'(t)\).

Given, 

\(\displaystyle p(t)=6t^{2}-7t+12\),

we can use the power rule which states, 

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\).

Applying this rule we can deduce that, 

\(\displaystyle v(t)=p'(t)=12t-7\).

Swapping in \(\displaystyle t=4\), we get 

\(\displaystyle v(4)=12(4)-7=48-7=41\).

Velocity is the first derivative of position, or \(\displaystyle v(t)=p'(t)\).

Given, 

\(\displaystyle p(t)=9t^{2}+14t+15\),

we can use the power rule which states, 

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\).

Applying the power rule we can deduce that, 

\(\displaystyle v(t)=p'(t)=18t+14\).

Swapping in \(\displaystyle t=5\), we get 

\(\displaystyle v(5)=18(5)+14=90+14=104\).

Since velocity is a rate change of position with respect to time, we need to take the derivative of the position function with respect to time:

\(\displaystyle v(t) = x'(t) = 2 \cdot (25 m/s^2 ) \cdot t\)

At 

\(\displaystyle t = 4 s\)

\(\displaystyle v(4 s) = (50 m/s^2) \cdot 4 s = 200 m/s\)