In order to find the distance travelled between  and , we need to take the definite integral of the velocity . 

Let's first define the definite integral as 

 for a continuous function  over a closed interval  with an antiderivative .

Using the inverse of the power rule 

with a constant  and , we can therefore determine that 

Completing the integral:

.

First, we must will figure out the particle's position with respect to time by integrating  with respect to . This gives us:

.

We know that  because we start at the origin. Thus,

.

This is a quadratic equation, so we know it has one local extrema. We can solve for this by settings its derivative equal to 0:

Solving this tells us that the extrema is at . Plugging this into  tells us that the particle is at  when it turns around. So we know that it starts at  moves to  and then turns around. After it turns around it goes to

.

This means it travels another  units.

Therefore, its total distance traveled is

 units.

We integrate the acceleartion function and use the inital velocity to find the velocity function:

We can now either choose to find the position function or take the definite integral of the velocity function over the given time interval to find the distance directly. Since we are not given the initial position, this is the natural approach.

The max height of the ball happens when the velocity is zero. Using our equation, 

 we can solve for the value of  for which the velocity is zero. 

This occurs at approximately . When , the height of the ball is .

This means the ball travels  meters up from the throw at the top of the building and then drops  to the ground, for a total of  travelled.

You cannot determine this without knowing how high the cannon is when it fires the t-shirt.

The t-shirt has no vertical velocity and initial height  so its vertical position is given by the function 

.

We can find the amount of time the t-shirt is airborn by setting .

We can make things a bit easier on ourselves by approximating

 and then set that expression equal to zero.

This gives us   seconds to reach the ground. Multiply this by the horizontal velocity  to get the horizontal distance travelled, which is  meters.

Her position as a function of time is just the integral of her , using the following rule,

 on each term we arrive at the position function which is,

 .

Between  and , she's displaced  units. 

This can be simplified to .

To solve this problem, it is important to understand that the derivative of position is velocity. Thus by integrating the velocity function, we will be able to obtain the posiiton function.  Because the problem asks us to find the distance the bird travels from  to , we integrate from 0 to 3.  The general formula for integration is

 where C can be any real number.  

Integrating the velocity function gives us 

, where C is any real number.  

However, the problem asks us the position change of the bird from   to , therefore the position of the bird is  at   and  at .  By subtracting, we get that the distance travelled by the bird during this time interval is .

We must first determine where the particle stops and turns around.

We do this by setting the first derivative, or velocity, equal to zero.

This yields .

Because  for  and

 for .

the distance travelled is

.

The total distance the puck travelled during the time interval t=1 to t=9 is given by the integral of the absolute value of the velocity equation over the given time interval: .

First, one must remove the absolute value from the integral by determining the intervals where the function is positive and negative. The critical values of the function (where it equals zero) are t=6 and t=8. On the interval [1,6) the function is positive, on the interval (6,8) the function is negative, and on the interval (8, 9] the function is positive. One can determine this by plugging in any number on the given interval into the function and seeing if the output is positive or negative. Now the original integral can be broken into three integrals: 

Note that the second integral has a negative sign in front. This is to account for the fact that the function is negative from t=6 to t=8, and we are solving for total distance, not displacement.

The integration of the function is:  , which comes from the rules  and .

To evaluate the function on each interval, one must plug the upper limit of integration into the function   and then subtract by the function when the bottom limit of integration is plugged in  .

Perform this operation for each of the three integrals, and the answer is .