Acceration is the derivative of velocity. This means that to go from acceleration to velocity we must take an integral. The integral of $\displaystyle x^n$ is $\displaystyle \frac{x^{n+1}}{n+1}$. Knowing this we can solve the problem.

$\displaystyle \int4t-7dt=2t^2-7t+v_{0}$

The constant at the end is the inital velocity so the final answer is 

$\displaystyle 2t^2-7t-2$

Velocity is the derivative of position. To find the velocity function all we have to do is take the fisrst derivative.

The derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$.

So the velocity function is

$\displaystyle v(t)=12t^2-9$

To find the velocity at a certain time, we must plug that time into the function.

$\displaystyle v(3)=12(3)^2-9=99$

When the question asks when the particle begins to move in an opposite direction, it's referring to a change in sign of the particle's velocity.

Velocity can be found as the time derivative of position.

For the position function

$\displaystyle x(t)=t^3-4t^2+4t+12$

The velocity is found to be

$\displaystyle v(t)=3t^2-8t+4$

The initial velocity is positive, since

$\displaystyle v(0)=0-0+4=4$

So we'll want to find where it first becomes negative; should this happen, it'll be right after the function equals zero. It may help to factor the equation:

$\displaystyle v(t)=(3t-2)(t-2)$

The velocity reaches zero at two times; $\displaystyle t=\frac{2}{3},2$, and is negative between them. It can be said then that the particle first changes directions at time

$\displaystyle t=\frac{2}{3}$

Begin by finding a function for velocity. Velocity is given as the integral of acceleration with respect to time.

For the acceleration function

$\displaystyle a(t)=2t+3$

The integral can be performed knowing that $\displaystyle \int at^n=\frac{at^{n+1}}{n+1}+C;(a:constant,n:constant \neq -1)$

The velocity function is then

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

The constant of integration is unknown as of this moment, so disregard it for the time being.

The distance traveled over an interval of time is given by the integral of the velocity function over said interval of time. For the interval of time $\displaystyle t=[0,5]$, the distance traveled can be calculated as follows:

$\displaystyle d=\int_0^5 (t^2+3t+C)dt$

$\displaystyle d=(\frac{t^3}{3}+\frac{3t^2}{2}+Ct)|_0^5$

$\displaystyle d=(\frac{125}{3}+\frac{75}{2}+5C)-(0+0+0)$

$\displaystyle d=\frac{125}{3}+\frac{75}{2}+5C$

$\displaystyle d=\frac{475}{6}+5C$

Now we're told that the distance traveled over this interval is $\displaystyle \frac{25}{6}$, so we can solve for the constant of integration:

$\displaystyle \frac{475}{6}+5C=\frac{25}{6}$

$\displaystyle 5C=-\frac{450}{6}$

$\displaystyle C=-15$

Putting this back into our velocity equation, we can find the initial velocity:

$\displaystyle v(0)=0^2+3(0)-15$

$\displaystyle v(0)=-15$

The velocity function can be found by integrating the acceleration function with respect to time.

For the acceleration function

$\displaystyle a(t)=4t+1$

Note that $\displaystyle \int at^n=\frac{at^{n+1}}{n+1}+C;(a,n:constants,n \neq -1)$

The velocity function is

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

To find this constant of integration, use what we're told in the problem: namely that the particle comes to rest (a zero velocity) at time $\displaystyle t=3$:

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

$\displaystyle C=-21$

So knowing the full velocity function

$\displaystyle v(t)=2t^2+t-21$

All that remains is to find the time when the velocity is $\displaystyle 34$:

$\displaystyle 2t^2+t-21=34$

$\displaystyle 2t^2+t-55=0$

$\displaystyle (2t+11)(t-5)=0$

There are two roots, $\displaystyle t=-5.5,5$. However, since a negative time is not possible, we elect the positive value:

$\displaystyle t=5$

Velocity can be found by taking the derivative of the position function. For the position function

$\displaystyle p(t)=at^2+bt+c$

Note that $\displaystyle \frac{d}{dt}at^n=nat^{n-1};(a,n:constants)$ and that the derivative a constant is zero.

The velocity function can thus be found to be

$\displaystyle v(t)=2at+b$

Currently the values of $\displaystyle a$ and $\displaystyle b$ are unknown, but we're given some information to find them. First, the initial velocity:

$\displaystyle v(0)=2a(0)+b=15$

$\displaystyle b=15$

$\displaystyle v(t)=2at+15$

Next, we're told the velocity at time $\displaystyle t=10$:

$\displaystyle v(10)=2a(10)+15=135$

$\displaystyle 20a=120$

$\displaystyle a=6$

$\displaystyle v(t)=2(6)t+15$

$\displaystyle v(t)=12t+15$

Now we can find the velocity at time $\displaystyle t=20$:

$\displaystyle v(20)=12(20)+15$

$\displaystyle v(20)=255$

If the cars distance traveled is given by the equation $\displaystyle d(t)=10t^2+65t$, we can see that the derivative $\displaystyle d'(t)=20t+65$ is another way of writing the equation for $\displaystyle v(t)$.  Because of this relation, we can see that at the start of the car's travel ($\displaystyle t=0$), the car was traveling at $\displaystyle 65\:\uptext{mph}$  $\displaystyle (d(0)=20(0)+65)$.  If we plug in a three for time (three seconds after the start) we can see that we come to our answer, $\displaystyle d(3)=20(3)+65.=125$.

Acceleration is the derivative of velocity. This makes velocity the integral of acceleration. The integral of $\displaystyle x^n$ is $\displaystyle \frac{x^{n+1}}{n+1}$. This makes the velocty function

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

We must include the $\displaystyle C$ because the integral is indefinite and there are no given values.

Velocity is the first derivative of position. The derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$. This means that the velocity function is given by 

$\displaystyle v(t)=8t-7$

We must plug in $\displaystyle t=2$ to get the velocity at that time.

$\displaystyle v(2)=8(2)-7=16-7=9$

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

 $\displaystyle p(t)=3^{\sqrt{t}}$

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

Derivative of an exponential: 

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

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

Using the above properties, the velocity function is

$\displaystyle v(t)=\frac{3^{\sqrt{t}}}{2\sqrt{t}}ln(3)$

$\displaystyle v(4)=\frac{3^{\sqrt{4}}}{2\sqrt{4}}ln(3)=2.47$