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:

If the particle is said to have returned to its starting point, that would mean that the total distance traveled is zero.

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

For the velocity function

Setting this equation equal to zero will provide the times that the particle returns to its starting point:

The equation is equal to zero at times , though we can disregard the trivial solution of .

The particle returns to its starting point for the first time at time

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

For the velocity function

A distance formula can be found as

This function is equal to zero at times  Discounting the trivial solution of , the particle first returns to its starting point at time

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

For velocity function

The distance over the interval of time  is

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

The velocity function is

The constant of integration can be found by using the initial condition given in the problem, that is :

Now, average velocity can be found by dividing distance traveled by total elapsed time. Furthermore, distance can be found over an interval of time by integrating the velocity function over said interval. Combining these two facts, average velocity can be given as:

In the case of this problem:

Note that the starting position was irrelevant.