The distance a particle travels during an interval of time  can be found by integrating the velocity function over that interval:

If velocity is given as:

Then the distance travelled in three seconds from the beginning of movement is:

Distance travelled can be found by integrating the velocity function over the interval of elapsed time :

If velocity is:

Then the distance requested in this problem is:

This integral can be found using the method of integration by parts:

Now, use the method of integration by parts once more, on the rightmost integral:

Now, with  terms on the left and right side of the equation, we can move the right-side term to the left to get:

Now, we can find at last the distance travelled:

To find the position of the object after a certain time, we first must find the position function of the rocket, then solve the position function at that given time.

The position function of an object moving with uniform acceleration is , where  is the inital position of the object,  is the inital velocity of the object and  is the acceleration of the object.

For this problem:

, , , 

After  seconds, the rocket will be at a height of .

Distance a particle travels can be found by integrating the velocity function over a specified interval with respect to time. For distance traveled from time a to time b:

For the velocity function

Since we're looking at a distance traveled from an initial position, we'll assume a starting time of zero . Furthermore, since we want to know the time that the position reaches its initial position, we'll want a total distance of travel of zero :

For a postive value of time, the particle will return to its original postion at a time of

Distance travelled can be found by integrating the velocity function over the specified interval of time :

For velocity 

The distance travelled over the interval specified is:

The distance traveled is the integral 

Note: The above integral can be evaluated using the Power Law for integrals, which says

Since we know both the initial velocity and constant deceleration, we can immediately determine the velocity function without integrating 

, where  is acceleration,  is time, and  is velocity at , which is also the initial velocity. 

To determine displacement , we take the definite integral of velocity over the first 3 seconds. To do this we need to utilize the power rule of integration.

By the power rule, we know that

, where are constants and is a variable.

We also know that for definite integrals, 

, where 

Joining these two rules of integration

The ball will have been displaced a total of , which in this case is equal to the distance travelled. 

To begin, find the velocity function by integrating acceleration with respect to time:

To find the integral we will need to apply the following rules.

Therfore we get,

.

Find the constant of integration by using the initial condition:

Now, find the distance traveled by integrating velocity over the interval specified:

Distance traveled over an interval of time  can be found by integrating the velocity function over that interval:

For the two velocity functions

There will be two distances traveled in the x and y directions:

The magnitude of distance traveled is then the root of the sum of squares

Distance traveled over an interval of time  can be found by integrating the velocity function with respect to time over that interval:

For the three velocity functions:

The distances traveled in the x, y, and z directions are therefore:

The magnitude of distance travelled is found then by taking the root of the sum of squares: