00:01
In this problem, we are given a velocity function for a particle.
00:05
We have that v as a function of t, the time is equal to t to the power of 3 minus 2 t squared.
00:15
We want to find the displacement and distance traveled by the particle during the time interval minus 1, 3.
00:23
Let's start by find the displacement.
00:26
The displacement s is defined as the integral a v of t over certain interval a to b.
00:43
Since we have the interval minus 1 .3, let's insert our upper and lower limits, and integrate our velocity function, t to the power 3 minus 2t squared d t.
01:00
Eviling this integral, we find that our displacement during this time interval is equal to t to the power of 4 over 4.
01:10
Minus 2 t to power 3 over 3.
01:15
Now you want to evaluate this between t is equal to minus 1 and t is equal to 3.
01:24
Evaluating our limit, we're going to find a displacement of 3 to the power 4 over 4 minus 2 times 3 to power 3 over 3 minus one quarter plus two thirds.
01:51
Being careful to add up all our fractions, we're going to find a displacement of four thirds.
02:03
Okay, so that's the result for displacement.
02:06
Now what's going to be our distance traveled? so for displacement, we only, for displacement we only care about the final point and initial point.
02:15
But for distance traveled, we need to be a little bit more careful.
02:18
Why is that? because of distance traveled, similarly corresponds to the integral of the velocity between a to v, but now we want to add an absolute value to our velocity, our velocity function.
02:40
The reason for that is because in case our protocol moves back in time, we want to add that distance.
02:46
In a displacement, we don't care if our particle moves back or four.
02:50
We only care about the initial and final position, but here we want to add these incremental changes.
02:59
So to integrate the absolute value, of our velocity we're going to do is we're going to actually look for regions where our velocity is negative and simply integrate by parts and when our velocity is negative we're going to add a minus sign so let's check when is v of t equal to zero once we find the zero then we can determine the regions where velocity is negative or positive so v of t is zero when t to the power of 3 minus 2 t square is equal to 0.
03:44
Factoring out that t square, we're going to have that t square times t minus 2 is equal to 0.
03:59
So this gives us two zeros.
04:02
T is equal to zero and t is equal to 3.
04:07
So let's check within which interval.
04:18
Is our velocity positive or negative...