(II) A car is moving with speed 16.0 m/s due south at one mome...

Question

(II) A car is moving with speed 16.0 m/s due south at one moment and 25.7 m/s due east 8.00 s later. Over this time interval, determine the magnitude and direction of (a) its average velocity, (b) its average acceleration. (c) What is its average speed? [Hint: Can you determine all these from the information given?]

Accepted Answer

Welcome back. Everyone in this problem. Commencing with an initial velocity of 12 m per second in the westward direction. A boat achieves a velocity of 18.5 m per second northward after a six second duration, we want to calculate the magnitude and direction of first, the average velocity second, the average acceleration through this interval. And third, the average speed for our answer choices. A says the average velocity is 22.1 m per 2nd, 57 degrees west of north. The second one says that the average acceleration is 1.08 m per second squared, 57 degrees west of north and that the speed is 30.6 m per second. B says the average velocity is 22.1 m per 2nd, 57 degrees west of north. The average acceleration is 1.08 m per second squared, 57 degrees north of west. And that the average speed cannot be determined. C says the average velocity and speed cannot be determined while the average acceleration is 3.67 m per second squared, 57 degrees north of west. And the D says the average velocity, average acceleration and average speed cannot be determined. Now, to better understand this problem, let's try to, let's try to draw a diagram to see what it says. Now, what are we saying here? Well, we're saying that our bot, OK. Our bot commences with an initial velocity of 12 m per second in the westward direction. So draw my little boat here. So first it goes 12 m per second in the western westward direction. And then it has a velocity of 18.5 m per second northward after six seconds. So afterwards, it goes northward. OK? And here we're talking about the velocity. And now if we notice here, we could make a, a form a triangle here. OK, where we have an angle theta or are the resultant of our velocity. Now, in this problem, first, we're trying to find the average velocity. And if we're trying to find the average velocity, let's think, what do we know about velocity? We recall that velocity is defined as the net displacement over time. In other words, a net displacement divided by the total time. Now, we know the time we know the average duration is six seconds, but we don't know the displacement. And there's nothing in the problem that tells us about the displacement. So since the displacement is not known, the average velocity cannot be determined. OK. So the average velocity cannot be determined. Now, with that in mind, let's try to figure out the average acceleration. Ok. So let's go down to 52 average acceleration. Now again, let's ask ourselves, what do we know about acceleration? We'll recall. OK. That average acceleration, average acceleration is going to be equal to a change in velocity divided by the change in time. OK. Because velocity acceleration just tells us how much velocity increases or decreases. Now, in this case, we know a change in velocity, we know that the boat started at 12 m per second and then ended at 18.5 m per second. So if we can create a vector for our velocity here and find the average velocity, then its magnitude and direction will give us the average acceleration. So first, let's do that. So we know that our change in velocity, we went from 18.5 m per second. OK? Or final velocity northward. So that's in the y direction or J as a vector. And initially, we started with 12 m per second westward. So that's minus 12 m per second. I OK. And we're dividing all of that, we're dividing our vector here or the change in velocity by our change in time, which is six seconds. Now, what do we know? We'll recall that time is a scalar. And when we're dividing a vector by a scalar, we can divide each component by that scalar. So that means then we could divide 18.5 m per second by six. So let me write that here. So we can divide that by six seconds and we can also divide 12 m per second by six seconds. Now, when we do that, we end up with our vector 18.5 divided by six being 3.083 J. Ok. And in the X direction 12 divided by six, that would have been minus two I and as a being acceleration that's meters per second squared. So what do we know now? Well, we know that our acceleration is 3.083 J minus two I. OK. So that's A and we have its X and Y components 3.083 for the Y component and negative two for the X component. Since we have the X and Y components, that means we can find the magnitude of our average acceleration because no recall. OK. Recall that the magnitude of a vector A is going to be equal to the square root of the sum of the squares of its X and Y components. So in this case, that means the magnitude of A or acceleration incidentally, it's still a I didn't do that on purpose guys. OK? But a magnitude of acceleration is going to be called to the square root of negative to I squared. OK. Plus 3.083 J squared. OK. And now when we put those components and when we put this expression into our calculator, we should get the magnitude of the acceleration to be equal to 3.6749 m per second squared or to the nearest 100th of a decimal place 3.67 m per second squared. OK. So now we have the average acceleration, the magnitude rather of the average acceleration. Next, we need to find the direction of our average acceleration. Again, recall that for a vector, the anger that the vector forms with a horizontal is going to be equal to the inverse tangent of its Y component divided by its X component. So that tells us then that our angle for our acceleration is going to be equal to the inverse tangent OK. Of its component, which we found to be 3.083 m per second squared divided by or X component, which we found to be 2 m per second squared. Now when we put this expression into our calculator, we get the angle to be 57 degrees. Now again, remember think about our direction, it's going in the negative X direction which is west and the positive Y direction which is north. So that is going to be 57 degrees north of west. OK. So that is going to be our average acceleration. Our average acceleration is 3.67 m per second squared at 57 degrees north of west. Finally, let's find our average speed. Now again, let's ask ourselves, what do we know about speed? We recall that speed. OK. Our average speed is equal to the distance that we've traveled divided by the total time. But in a way similar to our problem before, with the average velocity, we don't know the distance traveled. The problem doesn't say anything about that. So since the distance traveled is not known, the average speed cannot be determined. Thus, the average speed and velocity cannot be determined. And the average acceleration is 3.67 m per second squared, 57 degrees north of west. If we look back on our answer, choices, C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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