0:00
Hi there.
00:01
So for this problem, we're told that an object at reps begins rolling horizontally for a time that is equal to 5 .3 seconds and achieves a final velocity that is equal to 12 .2 meters per second.
00:23
Then the object then slows with an magnitude of acceleration that is equal to 2 .2.
00:31
0 .49 meters per second square for an in this for a distance that is equal to 18 meters.
00:43
So finally it travels backwards with the same velocity magnated as the end of the second segment for a for yes for a distance now with the same acceleration for a distance.
01:04
This, well, let's call the first distance one and this last distance two.
01:08
And that distance is going to be 4 .78 meters.
01:14
Yes.
01:15
So the question, the first question is, what is the acceleration of the first settlement? so we need to find the acceleration in this first segment.
01:23
So let's call the first settlement a, the second segment b, and this last segment c.
01:29
So for this first segment a, we will have an speed and a time.
01:34
So the acceleration is going to be positive and that acceleration is just the final speed that is 12 .2 meters per second, which is the final speed divided by the time that it takes to get to that speed, which is 5 .3 seconds.
01:51
So using our calculator, we obtained a value of 2 .3 meters per second square.
02:01
So that's a solution for the first question in here.
02:05
Now for the second question, we are told, what was the time to complete the second segment? now, in that case, so in this part for par b, we know that the initial speed is the final speed from before.
02:29
So that will be 12 .2 meters per second.
02:32
The final speed after this, and time of acceleration, this, yes, for the second part, is the value that we need to, well, find in order to find the s time.
02:45
And then, well, we are given that the acceleration in this case is at 2 .49 meters per second square.
02:57
So for that, first we are going to determine the final speed.
03:02
So for that we use an equation from kinematic that is that the final speed square is equal to the initial speed square.
03:10
And this, well, we are told that this is slowing down.
03:17
So that means minus in here, because the acceleration is a decaleration, minus two times acceleration times the distance div 1.
03:26
So now we just take the square root of both sides of this expression, just like this.
03:32
And now we substitute the value.
03:34
So that will be the square root of.
03:37
The initial speed to the square, so that will be 12 .2 meters per second to the square minus, two times the acceleration that we are given for this segment, which is 2 .49 meters per second to the square, these times the distance the one that we are given 18 meters per second, and 18 meters, just simply 18 meters.
04:01
And we take the square root of all of this.
04:03
So using our calculator, we obtain, a value off...