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 . 

The change in position from time 0 to time a is given by

.

By the information given to us in the question for the velocity function and the change in position, we can form the equation:

From here, we integrate the left side:

This equation can be solved for a using the quadratic equation:

Since time must be positive, the answer is 3.

To begin with, start by finding the velocity function, the integral of acceleration with respect to time:

, the integration of constant, may be found by knowing an initial condition. We are told the car starts at rest, meaning:

The distance the car travels in a given time is then given by an integral bounded by the start and finish times:

In this problem, we are given a distance and a starting time, so we must solve for the final time:

We are given that the trains deccelaration equation is  and that its initial velocity is .  With this information we can find the minimum distance the train needs to start breaking in order to avoid hitting the car.  We must realize that the integral of acceleration is velocity and the integral of velocity is position.  

Therefore we must take the integral of the acceleration formula in order to find the train's velocity equation.  

Using the general integral formula 

,

we find that the velocity formula for the train is .

Plugging in  to solve for , we find that , therefore the final velocity equation is .  

Now in order to find the position function, we must once again take the integral of the velocity equation to find the position equation.  

Using the general integral formula 

, we find that the position formula for the train is .  

Plugging in the 5 seconds the driver has to break, we find that the train must be at least 375 meters away in order to avoid hitting the car.

The distance an object has traveled after a given amount of time is the area under the speed curve. To find the area under s(t) between 0 and 1, the following definite integral must be evaluated:

Use u-subtitution to evaluate this integral. Let u = 2t. Then du = 2dt, and dt= du/2. We also need to find the new limits of integration. When t = 0, u = 2(0) = 0. When t = 1, u = 2(1) = 2. Using this information, we can rewrite the integral as: 

Since the antiderivative is:

Evaluating this antiderivative at the limits of integration leads to the final answer: