Begin by finding the velocity function by integrating the acceleration function.

We use  as the constant of integration since the function  is a velocity and at initially, at , the velocity is equals the constant of integration. 

At  seconds we are told the velocity is equal to . 
 

The average value of a function on a given interval  is given by the following function:

Now, let's simply input our values and function in:

The position function describing any object is the antiderivative of the velocity function (in other words, velocity is the derivative of position). 

So, we first integrate the velocity function:

The following rule was used for integration:

Now, to determine the constant of integration, we use our initial condition given,

Plugging this into our function, we get

Our final answer is

To find the position function from the acceleration function, we integrate the acceleration function to find the velocity function, and integrate again to get the position function:

The integral was found using the following rule:

To find the constant of integration, we use the initial velocity condition given:

Now, after replacing C with the known value, we integrate the velocity function to get the position function:

The same rule of integration was used as above.

We use the same procedure to solve for C, too, only this time using the initial position condition:

Our final answer is

We first must establish the following relationship

 and 

We now may note that 

 or 

Since 

We must plug in our initial condition 

Therefore our new velocity equation is 

We now may similarly integrate the velocity equation to find position.

Plugging in our second initial condition 

We find our final equation to be:

To find the velocity function, we integrate the acceleration function (the acceleration is the antiderivative of the velocity):

The rules of integration used were

, 

To solve for the integration constant, we plug in the given initial condition:

Our final answer is

To find the position function, we must integrate the velocity function, as velocity is the antiderivative of position:

The following rule of integration was used:

Finally, we use the initial condition to solve for the integration constant:

Our final answer is

In this problem, letting  denote the position of the particle and  denote the velocity, we know that . Integrating and working backwards we have,

Plugging in our initial condition, , we see immediately that .

Repeating the process again for , we find that 

Plugging in our initial condition,  (we started at the origin) we see that . This gives us a final equation

. The problem asks for  which is simply 

Find the integral which satisfies the specific conditions of 

To do this problem, we need to recall that integrals are also called anti-derivatives. This means that we can calculate integrals by reversing our integration rules.

Furthermore, to find the specific answer using initial conditions, we need to find our "c" at the end.

Thus, we can have the following rules.

Using these rules, we can find our answer:

Will become:

And so our anti-derivative is:

Now, let's find c. First set our above expression equal to y

Next, plug in   for y and t. Then solve for c

Looks a bit messy, but we can clean it up to get:

Now, to solve, simply replace c with 12.12

Begin by finding the velocity function by integrating the acceleration function.

We use  as the constant of integration since the function  is a velocity and at initially, at , the velocity is equals the constant of integration. 

At  seconds we are told the velocity is equal to .