To solve this question there are two steps involved. One, find the velocity function and two, substitute the specific time into the velocity function.

To get the equation for velocity, we take the derivative of the position equation.

Recall the rules of differentiation to solve this problem, specifically the power rule which states,

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

and the constant rule which states that the derivative of a constant is always zero.

Applying the power rule to each term of the position function will result in obtaining the velocity function.

\(\displaystyle \\v(t)=p'(t)\\p(t)=3t^2+6t-4\\p'(t)=2\cdot 3t^{2-1}+1\cdot 6t^{1-1}-0\\p'(t)=6t+6\)

\(\displaystyle v(t)=6t+6\).

Evaluating at \(\displaystyle t=6\), we get 

\(\displaystyle v(6)=6(6)+6=36+6=42\)

\(\displaystyle 42\,m/s\).

To solve this, we'll need to first differentiate the position function to get the velocity function, and then plug our specific time into that velocity funtion to get the velocity at that moment. Differentiating the position equation term-by-term, using the power rule, which states that

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

and the constant rule, which states that the derivative of a constant is 0, we get

\(\displaystyle \begin{align*}v(t)&= p'(t) \\ p(t)&=7t^2-15t+4 \\ p'(t) &= (2\cdot 7)t^{2-1}-15t^{1-1}+0 \\ p'(t)&=14t-15 \\v(t) &=14t-15\end{align*}\)

Evaluating at  \(\displaystyle t=3\) gives us 

\(\displaystyle \begin{align*} v(3)&= 14(3)-15 \\ &=42-15\\&=27\,m/s \end{align*}\)

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: \(\displaystyle p'(t)=v(t)\)

In this case, the position function is: \(\displaystyle p(t)= 2(\frac{1}{2}t+3)^2\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= t+6\)

Then, plug \(\displaystyle t= -5\) into the velocity function: \(\displaystyle v(-5)= {\color{Blue}-5 }+6\)

Therefore, the answer is: \(\displaystyle 1\)

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: \(\displaystyle p'(t)=v(t)\)

In this case, the position function is: \(\displaystyle p(t)= -ln(2t^2 + 5t)\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= - \frac{4t+5}{2t^2 + 5t}\)

Then, plug \(\displaystyle t= 1\) into the velocity function: \(\displaystyle v(1)= - \frac{4({\color{Blue} 1})+5}{2({\color{Blue} 1})^2 + 5({\color{Blue} 1})}\)

Therefore, the answer is: \(\displaystyle -\frac{9}{7}\)

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: \(\displaystyle p'(t)=v(t)\)

In this case, the position function is: \(\displaystyle p(t)= \cos(2t +e^t)\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= -\sin(2t +e^t) (2+e^t)\)

Then, plug \(\displaystyle t= 0\) into the velocity function: \(\displaystyle v(0)= -\sin(2({\color{Blue}0 }) +e^{{\color{Blue} 0}}) (2+e^{{\color{Blue} 0}})\)

Therefore, the answer is: \(\displaystyle -2.5\)

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: \(\displaystyle p'(t)=v(t)\)

In this case, the position function is: \(\displaystyle p(t)= \cos(2t +e^t)\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= -\sin(2t +e^t) (2+e^t)\)

Then, plug \(\displaystyle t= 0\) into the velocity function: \(\displaystyle v(0)= -\sin(2({\color{Blue}0 }) +e^{{\color{Blue} 0}}) (2+e^{{\color{Blue} 0}})\)

Therefore, the answer is: \(\displaystyle -2.5\)

To solve this problem, you must first find the velocity function. To do that you, you need to find the derivative of the position function. Remember, when taking the derivative, multiply the exponent by the coefficient in front of the x and then subtract 1 from the exponent. Therefore, the velocity function is: \(\displaystyle v(t)=4x-1\). Then, set that expression equal to 0 to get your x: \(\displaystyle 4x-1=0; x=\frac{1}{4}\)

To answer this problem one must know that velocity is the first derivative of position therefore \(\displaystyle r'(t)=v(t)\).

The velocity function for our particle is the derivative of our given position function:

\(\displaystyle r(t)=Cos(t)+sin(t)-5\)

\(\displaystyle v(t)=r'(t)=-Sin(t)+Cos(t)\)

Evaluated at point \(\displaystyle t=\pi/4\):

\(\displaystyle v(\pi/4)=-Cos(\pi/4)+Sin(\pi/4)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=0\)

Therefore, at time \(\displaystyle t=\pi/4\), the velocity is 0.

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: \(\displaystyle p'(t)=v(t)\)

In this case, the position function is: \(\displaystyle p(t)= e^{2t} + \frac{1}{ln(2t)}\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= 2e^{2t} - \frac{1}{t ln^2(2t)}\)

Then, plug \(\displaystyle t= 1\) into the velocity function: \(\displaystyle v(t)= 2e^{2({\color{Blue} 1})} - \frac{1}{({\color{Blue} 1}) ln^2(2({\color{Blue} 1}))}\)

Therefore, the answer is: 

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: \(\displaystyle p'(t)=v(t)\)

In this case, the position function is: \(\displaystyle p(t)= \sin(3t^2)\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)=6t \cos(3t^2)\)

Then, plug \(\displaystyle t= 2\) into the velocity function: \(\displaystyle v({\color{Blue} 2})=6({\color{Blue} 2}) \cos(3({\color{Blue} 2})^2)\)

Therefore, the answer is: \(\displaystyle 10.1\)