00:01
In this question, we are given a particle's velocity function and we are asked to find its displacement and distance over the given interval.
00:09
To find the particle's displacement, we need to calculate the integral from 0 to 5 of the velocity function.
00:26
The integral of 3 t minus 8 equals to 3 times t squared over 2 minus 8 t in substitution from 0 to 5.
00:38
After plugging in limits, we are going to get 3 halves times 2.
00:43
5 squared minus 8 times 5 and after plugging in 0 we're just going to get 0 that's going to be 3 over 2 times 25 minus 40 3 over 2 times 25 equals to 75 over 2 and we can rewrite 40 as 80 over 2 and 75 over 2 minus 80 over 2 equals to negative 5 over 2 so this is the object's displacement this means the object moved 5 half meters to the left from its initial position.
01:28
Now to find the object's distance, we need to calculate the integral from 0 to 5 of the absolute value of v of t.
01:38
Now you will see the difference.
01:44
So we need to integrate the absolute value of 3t minus 8.
01:58
What we are going to do now is draw the number line.
02:06
We also need to find the point where 3t minus 8 equals to 0.
02:16
When 3 t equals to 8 or t equals to 8 over 3 so there is a point t equals to 8 over 3 now when t is greater than 8 over 3 well 3 t minus 8 is greater than 0 if 3 t is greater than 8 over 3 so over when t is greater than 8 over 3 the particle's velocity is positive when t is less than 8 over 3, the particle's velocity is negative.
03:10
Recall that 3t minus 8 represents the particle's velocity, right? this is v of t...