00:01
In this question, given the particle's velocity function, we are asked to find its displacement over the interval 05, its total distance, and the average velocity over the same interval.
00:12
To find the displacement, we just need to calculate the integral of v of t from 0 to 5.
00:26
So we need to integrate 3t squared minus 12t.
00:40
We are going to get 3t cube over 3 minus 12t squared over 2.
00:49
In substitution from 0 to 5, which simplifies to t -cube minus 6t squared, in substitution from 0 to 5.
01:07
And after plugging in the limits, we are going to get 1 .25 minus 6 times 25, and this equals to 1 .25 minus 150 equals to negative 25.
01:24
This is a particle displacement over the interval.
01:32
Next, let's find the total distance.
01:33
To find the total distance, we need to integrate the particles, the absolute value of the velocity function.
01:50
So we need to integrate the absolute value of 3t squared minus 12t.
02:04
To do that, to calculate this interval, we need to know the points where it's positive and where it's negative.
02:14
We can rewrite it as 3t times t minus 4, right? 3t times t minus 4 and on the number line and this function becomes 0 at t equals 0 and t equals 4 and these two points they split the number line into three sub -intervals from 4 to infinity 3 t times t minus 4 when t is greater than 4 t minus 4 is positive and t is also positive right and this means that the whole expression is positive from 4 to infinity.
03:04
So v of t is positive from 4 to infinity...