(II) A hiker follows a winding trail for 5.5 hours while climb...

Question

(II) A hiker follows a winding trail for 5.5 hours while climbing a mountain. The distance along the trail is 11.5 km and the summit is 850 m above and 8.0 km due north of the starting point. What are the average speed and the magnitude and direction of the average velocity vector?

Accepted Answer

Hey, everyone in this problem, a trekker navigates a meandering path for 5.6 hours to ascend a peak. The journey spans 11.6 kilometers culminating at a summit 860 m high positioned nine kilometers due north of the commencement point. We're asked to assess the trekkers mean speed and the average velocity vector magnitude and orientation. We're given four answer choices A through D each of them containing different values for these things that we need to find. We're gonna come back to these answer choices as we work through the problem. So let's start with the first thing we're asked to find and the first thing we're asked to find is the mean speed. OK? So when we're thinking about the mean speed, all we need to be thinking about is how far we're going and how long it's taking. OK. So the means speed we call is gonna be equal to the distance that we travel divided by the time it takes. OK? And when we're talking about speed, this is distance, not displacement. These are not the difference in the initial and final positions, but the actual distance we travel now in this problem, we're told that we have this journey that spans 11.6 kilometers. And so the distance traveled is 11.6 kilometers. We have 11.6 kilometers and this takes 5.6 hours. So divided by 5.6 hours. And when we work this out, we get that the average speed or the mean speed, it's gonna be 2.07 kilometers per hour. So that is the answer to the first part of this question. We can take a look at our answer choices and see how we're doing. OK. So option AC and D all have this average speed that we found of 2.07 kilometers per hour. Option B has 6.07 kilometers per hour. That is not what we found. So we can eliminate that as an option. We're down to three answer choices and we're gonna move on to the rest of the question. Now, the rest of the question is asking about the average velocity vector. So we want the magnitude and we want the direction. So let's think about this velocity vector and we're gonna think about this hill that we're climbing. OK. So we're trekking up a hill. Now, this is not drawn in the correct direction. OK. We're traveling straight north. You can imagine that this triangle would be turned so that you're really only looking at the hypotenuse it's pointing up in front of you. We're gonna turn it on its side. So we can take a look at the angle. And so we're checking up some angle, we know that this peak is nine kilometers from our commencement point. So along the ground and this horizontal distance is nine kilometers and we know that the height of that peak is 860 m, ok? So the height is 860 m or we can write this as 0.86 kilometers. OK? So our triangle is not drawn to scale at all, but we have these two sides and that's gonna allow us to find this angle theta and it's also gonna allow us to find the displacement delta X. OK. So finding that displacement will allow us to calculate the magnitude of the average velocity vector. So let's start there. We're gonna start by calculating delta X. So delta X, hey, this is a right hand triangle. So we're gonna use Pythagorean theorem. Delta X squared. It's just gonna be equal to nine kilometers squared, 10.86 kilometers squared. When we work this out, we get that delta X is going to be equal to 9.041 kilometers approximately nice. So we've squared everything on the right hand side and then we've taken the square root. OK. So 9.041 kilometers is the displacement. OK? When we're thinking about average velocity, OK. The average velocity is going to be that displacement. Delta X divided by the time T, OK. When we did the average speed, it was just the distance we took. Now we're looking at the displacement so we can substitute that in 9.0 or one kilometers divided by the time 5.6 hours. And we get the average velocity as 1.61 45 kilometers per hour. OK? We have the average speed. We have the average velocity. We can take a look at our answer choices again, hey, we have the magnitude, that's what we've calculated. But we also need the direction and we need the vector itself. OK. So this direction, well, this is gonna be our theta value. OK. So we can go ahead and calculate the theta value. We can find the using tangent because we're gonna relate the opposite and adjacent sides. Those are the sides that we knew that we were given in the problem and that we have without any round off error. So tangent of that angle theta is going to be equal to the opposite side, 0.86 kilometers divided by the adjacent side nine kilometers. If we take the inverse tangent on both sides, we get that data will be equal to about 5.458 degrees. So we have to think about what this direction actually means. Remember that we've turned our picture kind of on its side. This angle is actually the elevation. OK. So the peak is pointed straight north and this angle is the angle of elevation from the ground upwards. Mhm. OK. So that's our angle. So that's the direction we were wanting to find that there's a direction above the horizontal that's still pointing to north. The last thing we need to do is find the actual velocity vector, but we can do that now that we have the angle and the average velocity. So if we go give ourselves some more room and we draw another triangle, hey, this triangle is going to relate the speed or the velocities. OK. So we know that the average velocity that hypotenuse is 1.61 or five kilometers per hour, we know that the angle is 5.458 degrees. And we wanna find the two components, the VJ component and the VK component. Now we know that this is the VJ component because this is actually pointing in the northward direction. OK. So when we typically draw our directions that's pointing upwards, that's the J component. And then K is pointing up out of the page. OK. That's the elevation, that's our K component like the Z component. OK. All right. So we just need to do some more triangle math here. If we're looking for side VJ, we know it's gonna be related through cosine because it is the adjacent side. So VJ is gonna be equal to that hypo 1.61 four or five kilometers per hour multiplied by cosine of 5.458 degrees. This gives us AJ component of our velocity of 1.60718 kilometers per hour. And we can do the same for the K component. And we take that hypotenuse 1.6145 kilometers per hour. And we multiply it this time by sign of that angle 5.45 eight degrees. And this gives us a K component that is equal to about 0.15 36 kilometers per hour. OK. So this angle is really, really small. That means that the elevation is quite small. And so our velocity is way more pronounced in the Y direction J component than in the cater. OK. All right. So the last thing we have to do, put this together into a vector. So we have that our average velocity vector is going to be this J component 1.60718 kilometers per hour J hot plus the K component 0.1536 kilometers per hour K hat. And now we have found everything that the question was asking for and we have our speed, our velocity, our direction and the velocity vector. Let's go back to our answer choices. We had already narrowed it down and eliminated option A or sorry. Option B, we had eliminated option B. If we take a look at the remaining answers, we can see that the correct answer is going to be option A it matches for all of those values. So we have the average speed 2.07 kilometers per hour. The average velocity vector 1.61 J hat or 0.15 k hat kilometers per hour, the magnitude of the velocity vector 1.61 kilometers. And finally, the direction is 5.4 degrees above the horizontal pointing due north. Thanks everyone for watching. I hope this video helped see you in the next one.

(II) A hiker follows a winding trail for 5.5 hours while climbing a mountain. The distance along the trail is 11.5 km and the summit is 850 m above and 8.0 km due north of the starting point. What are the average speed and the magnitude and direction of the average velocity vector?

Key Concepts

Average Speed

Average Velocity

Displacement

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