The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$\displaystyle y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$\displaystyle p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of $\displaystyle t$ and divide by the number of steps to be used:

$\displaystyle \Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told $\displaystyle v(t)=cos^3(\frac{\pi}{6}t!)$ $\displaystyle p(0)=5$

Knowing this, we may take the steps to estimate our function value at our desired $\displaystyle t$ value:

$\displaystyle \Delta t = \frac{3-0}{3}=1$

$\displaystyle p_0=5;t_0=0$

$\displaystyle p_1=5+(1)cos^3(\frac{\pi}{6}(0)!)=5+\frac{3\sqrt{3}}{8}$

$\displaystyle p_2=5+\frac{3\sqrt{3}}{8}+(1)cos^3(\frac{\pi}{6}(1)!)=5+\frac{3\sqrt{3}}{4}$

$\displaystyle p_3=5+\frac{3\sqrt{3}}{4}+(1)cos^3(\frac{\pi}{6}(2)!)=\frac{41}{8}+\frac{3\sqrt{3}}{4}=\frac{41+6\sqrt{3}}{8}$

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: $\displaystyle p'(t)= v(t)$

In this case, the position function is: $\displaystyle p(t)= \frac{2}{3} t^3 + ln(t^2)+ 5$

The velocity function is found by taking the derivative of the position function: $\displaystyle v(t)= 2 t^2 + \frac{2}{t}$

Finally, to find the velocity, substitute $\displaystyle t=1$ into the velocity function: $\displaystyle v(1)= 2 (1)^2 + \frac{2}{(1)}$

Therefore, the answer is: $\displaystyle 4$

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: $\displaystyle p(t)= ln(2t)$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t)= \frac{1}{t}$

Then, plug $\displaystyle t= 1$ into the velocity function: $\displaystyle v(t)= \frac{1}{{\color{Blue} 1}}$

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: $\displaystyle p(t) = (t-4)^3$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = 3(t-4)^2$

Then, plug $\displaystyle t= 2$ into the velocity function: $\displaystyle v(t) = 3({\color{Blue} 2}-4)^2$

Therefore, the answer is: $\displaystyle 12$

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: $\displaystyle p(t)= e^{3t^2+4t+1}$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t)= e^{3t^2+4t+1} (6t+4)$

Then, plug $\displaystyle t= 0$ into the velocity function: $\displaystyle v(t)= e^{3({\color{Blue} 0})^2+4({\color{Blue} 0})+1} (6({\color{Blue} 0})+4)$

Therefore, the answer is: $\displaystyle 10.9$

We'll need to work backwards to solve this problem. Let's start with the information we're given.

Let's just say that westward motion is positive and eastward motion is negative. We could flip it around, but it would give us the same result, albeit with all the signs flipped.

Since the motorcycle starts out at rest relative to the train, and the train is moving at a constant speed, the initial velocity is given as
$\displaystyle v(0)=-72\,km/hr$.

The motorcycle accelerates westward, so
$\displaystyle a(t)=2\,m/s^2$.

At this point, we have a problem. The acceleration is in meters per second, but the initial velocity is in kilometers per hour. To convert the initial speed into meters per second, we need to multiply as follows:

$\displaystyle \frac{72\,km}{1\,hr}\cdot \frac{1000\,m}{1\,km}\cdot\frac{1\,hr}{60\,min}\cdot\frac{\,min}{60\,s}=20\,m/s$

Initial and final time are given as:
$\displaystyle t_i=0\,s$
$\displaystyle t_f=10\,s$

The aceleration is the rate of change in the velocity, so to find the velocity function, we will need to integrate and use initial conditions.

$\displaystyle \int a(t) dt = \int 2dt$

To integrate, we use the following rule:

$\displaystyle \int adt = ax+C$

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

$\displaystyle \int a(t)dt=\int 2dt =2t+C$.

To solve for C, we'll use out initial condition, $\displaystyle v(0)=-20$.

$\displaystyle \\ v(0)=2(0)+C=-20 \\ C=-20 \\ v(t)=2t-20$

We then need to evaluate the equation $\displaystyle v=2t-20$ for $\displaystyle t=10$, which gives us our answer:

$\displaystyle \begin{align*}v(10) &=2(10)-20 \\ &= 20-20 \\ &= 0\,m/s \end{align*}$

At this instand, Mr. Bond is not moving relative to the ground.

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: $\displaystyle p(t)= 2t^4 + \cos(t^2) - ln(2t)$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t)= 8t^3 - 2t\sin(t^2) - \frac{1}{t}$

Then, plug $\displaystyle t= 1$ into the velocity function: $\displaystyle v(1)= 8({\color{Blue} 1})^3 - 2({\color{Blue} 1})\sin({\color{Blue} 1}^2) - \frac{1}{{\color{Blue} 1}}$

Therefore, the answer is: $\displaystyle 5.3$

Differentiating the equation gives us the equation for velocity, $\displaystyle v(t)=8t+5$, which, evaluated at $\displaystyle t=2$, gives us $\displaystyle 21\,m/s$.

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

$\displaystyle \begin{align*}v(t)&= p'(t) \\ p(t)&=4t^2+5t-16 \\ p'(t) &= (2\cdot 4)t^{2-1}+5t^{1-1}+0 \\ p'(t)&=8t+5 \\v(t) &=8t+5\end{align*}$

Evaluating at  $\displaystyle t=2$ gives us 

$\displaystyle \begin{align*} v(3)&= 8(2)+5 \\ &=16+5\\&=21\,m/s \end{align*}$

Velocity is the rate of change of position, so to find the velocity, we'll need to differentiate this equation, by applying the product rule

$\displaystyle (fg)'=(f'g)+(fg')$

This gives us:

 $\displaystyle \begin{align*}v(t)&= (\sin t \cos t)'\\&= (\cos t \cdot \cos t)+(\sin t\cdot -\sin t)\\&= \cos^2 t-\sin^2 t\end{align*}$

To solve this particular problem we will need to work backwards. First identify all known functions and values.

Velocity is given as,

$\displaystyle v(0)=50\ m/s$.

Acceleration is given as,

$\displaystyle a(t)=-9.8\ m/s^2$.

The initial time and final time are given as,

$\displaystyle t_i=0$

$\displaystyle t_f=5$.

The acceleration is the rate of change in the velocity, and in this case, it acts against the cannonball's initial motion.

To find the velocity function we will need to integrate the acceleration function and use initial conditions.

$\displaystyle \int a(t)dt=\int -9.8dt$

To integrate use the following rule of integration,

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}$.

Thus,

$\displaystyle \int a(t)dt=\int -9.8dt=-9.8t+C$

Remember when integrating we must include the unknown constant C. To solve for C we will use the initial condition of $\displaystyle v(0)=50\ m/s$.

$\displaystyle \\v(t)=-9.8+C\rightarrow v(0)=50\\50=-9.8(0)+C \rightarrow C=50\\v(t)=-9.8t+50$

So we need to evaluate the equation $\displaystyle v(t)=-9.8t+50$ at $\displaystyle t=5$, which gives us a velocity of 

$\displaystyle \\v(5)=-9.8(5)+50\\v(5)=-49+50$

$\displaystyle v(5)=1\,m/s$ upward.