Question

The velocity of a particle moving along the x-axis is given by the function v(t) = 3t^2 - 6t, where t is measured in seconds and v(t) is measured in meters per second. Find the net change in the position of the particle on the interval [1, 3] and determine the total distance the particle traveled on this interval.

Answer

To find the net change in the position of the particle on the interval [1, 3], we need to integrate the velocity function v(t) from 1 to 3:

Net change = ∫(1 to 3) v(t) dt

First, let's find the antiderivative of v(t):

∫(3t^2 - 6t) dt = t^3 - 3t^2 + C

Now, we can evaluate the definite integral:

Net change = [t^3 - 3t^2] from 1 to 3 Net change = [3^3 - 33^2] - [1^3 - 31^2] Net change = [27 - 27] - [1 - 3] Net change = 0 - (-2) Net change = 2 meters

So, the net change in the position of the particle on the interval [1, 3] is 2 meters.

Next, let's determine the total distance the particle traveled on this interval. This can be found by taking the absolute value of the net change and integrating the absolute value of the velocity function over the interval [1, 3]:

Total distance = ∫(1 to 3) |v(t)| dt

First, let's find the absolute value of the velocity function:

|3t^2 - 6t| = 3t^2 - 6t for t in [1, 3]

Now let's integrate the absolute value of the velocity function:

Total distance = ∫(1 to 3) (3t^2 - 6t) dt

We already have the antiderivative of 3t^2 - 6t, so we can proceed with the integration:

Total distance = t^3 - 3t^2 from 1 to 3 Total distance = [3^3 - 33^2] - [1^3 - 31^2] Total distance = [27 - 27] - [1 - 3] Total distance = 0 - (-2) Total distance = 2 meters

So, the total distance the particle traveled on the interval [1, 3] is 2 meters.