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 

$\displaystyle t = 4 s$

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