We know 

$\displaystyle p(t)=t^3-2t^2-6t-9$ and using the power rule to differentiate the position funtion we can find the velocity function.

The power rule states, $\displaystyle (x^n)'=nx^{n-1}$.

Therefore, the velocity function becomes,

$\displaystyle v(t)=p'(t)=(t^3-2t^2-6t-9)'$

$\displaystyle v(t)=3t^{3-1}-2(2t^{2-1})-6t^{1-1}-0$

$\displaystyle v(t)=3t^2-4t-6$.

We also know that $\displaystyle v(t)=0$ at $\displaystyle -0.897$ and $\displaystyle 2.230\ seconds$. Because $\displaystyle t< 0$ makes no sense, we can eliminate anything prior to that. Now we have $\displaystyle 2$ intervals. 

$\displaystyle (0,2.230),(2.230,\infty)$

Choosing a value from each interval, we can determine if the velocity is positive (the particle is moving to the right) or negative (meaning that the particle is moving to the left). You can also do this with a graph.

$\displaystyle v(1)=3-4-6=-7< 0$

$\displaystyle v(3)=27-12-6=9>0$

Thus the particle is moving to the right on the interval $\displaystyle (2.230,\infty)$ or anytime after $\displaystyle 2.230\ seconds$ have passed.

To find velocity, simply differentiate the position function and then plug in $\displaystyle t=3$. Thus,

$\displaystyle v(t)=p'(t)=6t-5$

$\displaystyle v(3)=6(3)-5=13$

To solve, simply differentiate using the power rule and then plug in $\displaystyle t=1$.

Recall the power rule:

$\displaystyle f(x)=x^n$

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

Thus,

$\displaystyle v(t)=p'(t)=6t$

$\displaystyle v(t)=6(1)=6$

Acceleration is the time derivative of velocity; conversely, velocity can be found by integrating a known acceleration function with respect for time:

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

For the acceleration function

 $\displaystyle a(t)=6t^2+14t+3$

The velocity function is:

$\displaystyle v(t)=2t^3+7t^2+3t+C$

Now, notice the inclusion of the constant of integration of $\displaystyle C$. Since in the process of derivation all constants are eliminated, it stands to reason that upon integrating back to an original function, we must consider that this original function had a constant included. To solve for this value, use the condition given in the problem for a point in time:

 $\displaystyle v(1)=2(1)^3+7(1)^2+3(1)+C=11$

$\displaystyle C=-1$

$\displaystyle v(t)=2t^3+7t^2+3t-1$

With the knowledge of the full velocity function, the value of the function can be found at time $\displaystyle t=2$

$\displaystyle v(2)=2(2)^3+7(2)^2+3(2)-1$

$\displaystyle v(2)=49$

Acceleration is the time derivative of velocity; conversely, velocity can be found by integrating a known acceleration function with respect for time:

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

For the acceleration function

$\displaystyle a(t)=4te^{t^2}+3t^2$

The velocity function can be found via knowledge of the following derivative properties:

Derivative of an exponential: 

$\displaystyle d[e^u]=e^udu$

The velocity function is:

 $\displaystyle v(t)=2e^{t^2}+t^3+C$

Now, notice the inclusion of the constant of integration of $\displaystyle C$. Since in the process of derivation all constants are eliminated, it stands to reason that upon integrating back to an original function, we must consider that this original function had a constant included. To solve for this value, use the condition given in the problem for a point in time. We're told that the particle begins at rest, so at time $\displaystyle t=0$:

 $\displaystyle v(0)=2e^{0^2}+0^3+C=0$

$\displaystyle C=-2$

$\displaystyle v(t)=2e^{t^2}+t^3-2$

With the knowledge of the full velocity function, the value of the function can be found at time $\displaystyle t=1$

$\displaystyle v(1)=2e^{1^2}+1^3-2$

$\displaystyle v(1)=2e-1$

Velocity of a particle can be found by taking the derivative of the position function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of position with respect to time, we are evaluating how position changes over time; i.e velocity!

$\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$\displaystyle p(t)=\tan(t^2+3t+4)$

We'll need to make use of the following derivative rule(s):

$\displaystyle d[\tan(u)]=\frac{du}{\cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$\displaystyle v(t)=\frac{2t+3}{\cos^2(t^2+3t+4)}$

At time $\displaystyle t=2$

$\displaystyle v(2)=\frac{2(2)+3}{\cos^2(2^2+3(2)+4)}$

$\displaystyle v(2)=374.39$

Velocity of a particle can be found by taking the derivative of the position function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of position with respect to time, we are evaluating how position changes over time; i.e velocity!

$\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$\displaystyle p(t)=\tan(4^{t})$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

$\displaystyle d[a^u]=a^udu\ln(a)$

Trigonometric derivative: 

$\displaystyle d[\tan(u)]=\frac{du}{\cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$\displaystyle v(t)=\frac{4^{t}\ln(4)}{\cos^2(4^{t})}$

At time $\displaystyle t=3$

$\displaystyle v(3)=\frac{4^{3}\ln(4)}{\cos^2(4^{3})}$

$\displaystyle v(3)=577.80$

Velocity of a particle can be found by taking the derivative of the position function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of position with respect to time, we are evaluating how position changes over time; i.e velocity!

$\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$\displaystyle p(t)=e^{\cos(\ln(2t))}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

$\displaystyle d[e^u]=e^udu$

Derivative of a natural log: 

$\displaystyle d[ln(u)]=\frac{du}{u}$

Trigonometric derivative: 

$\displaystyle d[\cos(u)]=-\sin(u)du$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$\displaystyle v(t)=\frac{-\sin(\ln(2t))e^{\cos(\ln(2t))}}{t}$

At time $\displaystyle t=1.2$

$\displaystyle v(1.2)=\frac{-\sin(\ln(2(1.2)))e^{\cos(\ln(2(1.2)))}}{1.2}$

$\displaystyle v(1.2)=-1.21$

Acceleration is the time derivative of velocity; conversely, velocity can be found by integrating a known acceleration function with respect for time:

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

For the acceleration function

$\displaystyle a(t)=3t\cos(t^2)+4t^2\sin(t^3)$

The velocity function can be found via knowledge of the following derivative properties:

Trigonometric derivative: 

$\displaystyle d[\sin(u)]=\cos(u)du$

$\displaystyle d[cos(u)]=-\sin(u)du$

The velocity function is:

$\displaystyle v(t)=\frac{3\sin(t^2)}{2}-\frac{4\cos(t^3)}{3}+C$

Now, notice the inclusion of the constant of integration of $\displaystyle C$. Since in the process of derivation all constants are eliminated, it stands to reason that upon integrating back to an original function, we must consider that this original function had a constant included. To solve for this value, use the condition given in the problem for a point in time:

$\displaystyle v(0)=\frac{3\sin(0^2)}{2}-\frac{4\cos(0^3)}{3}+C=0$

$\displaystyle C=\frac{4}{3}$

$\displaystyle v(t)=\frac{3\sin(t^2)}{2}-\frac{4\cos(t^3)}{3}+\frac{4}{3}$

With the knowledge of the full velocity function, the value of the function can be found at time $\displaystyle t=3$

$\displaystyle v(3)=\frac{3\sin(3^2)}{2}-\frac{4\cos(3^3)}{3}+\frac{4}{3}$

$\displaystyle v(3)=2.34$

The velocity function is given by the first derivative of the position function:

$\displaystyle v(t)=s'(t)=e^t\sin(t^3)+3t^2e^t\cos(t^3)-\frac{3}{t^4}$

The derivative was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$