A hiker travels 1.50 km north and turns to a heading of 20.0^∘ north of west, traveling another

step 1 Once again, welcome to a new problem. This time we're dealing with vector quantity and vector qu

A hiker travels 1.50 km north and turns to a heading of...

Key Concepts

Coordinate Systems

Using a coordinate system helps in representing directional movements in a structured way. Assigning a coordinate system (commonly with east as positive x and north as positive y) allows the translation of directional headings into numerical components for easier analysis.

Trigonometry in Displacements

Trigonometry is used to relate the angles of movement with their corresponding horizontal and vertical components. Functions such as sine and cosine are applied to determine these components when the movement is given at an angle relative to the primary directions.

Vector Decomposition

This concept involves breaking a vector into its horizontal and vertical components using trigonometric functions like sine and cosine. It is essential for solving problems where movements are not aligned with the coordinate axes, allowing each segment of displacement to be expressed in terms of its coordinates.

Vector Addition

Vector addition is the process of combining multiple vectors to find a resultant vector. By summing the individual components of each vector along the relevant axes, the net displacement from the starting point can be determined.

Transcript

00:07 Once again, welcome to a new problem.

00:10 This time we're dealing with vector quantity and vector quantities a little bit different from scalar quantities.

00:19 Vector quantities have magnitude, they have magnitude and directions.

00:26 So those are the two things that vector quantities are.

00:30 For example, displacement is a vector quantity because it has two components.

00:37 It has distance and also it has the direction and the direction in terms of distance can be angular directions or it could also be in terms of you know left right east west north south and then positive and negative directions.

01:07 So those are things that you get to see when dealing with displacement.

01:14 And for example, if i had an object at this point and then moved towards the right, a distance, say, 20 meters, then my displacement would be 20 meters to the right.

01:28 If i make a u -turn in terms of angular direction, that would be a spin of 180 degrees, and then go back to the same spot it means that my displacement my displacement would be zero and the reason for that is because you're still back at the same position as you studied off before so coming back to this problem we have a case involving hiking so we have a case involving hiking so let's assume that let's assume that we have a hiker, we have someone hiking and they are obviously going to move in a specific direction.

02:21 So we start at this point and they travel a distance, they travel a distance d1, the travel a distance, d1 so that's the distance they travel and they're going north and the distance they traveled is 1 .5 kilometers and then they're gonna change direction and now they're gonna move in the northwest direction at an angle at an angle that's equivalent to 20 degrees so this angle is 20 degrees.

03:10 They're going north and remember in terms of directions this is north this is south this is west this is east so this direction would be north -west and they're traveling a second distance that's equivalent to 1 .5 kilometers so they traveled 1 .5 kilometers a distance in the northwest direction and then they're going to go back up again northwards and northwards again they travel a distance equivalent to 1 .5 kilometers so that's that's their that's their direction in terms of direction we could also have the x direction and the y direction so in this particular problem is determine determine determine the distance and i'm going to call this point a and this point b.

04:19 So the gap between these two, this is the distance you're looking for a to b.

04:28 So this distance is what you want to get.

04:33 Determine the distance a .b and the direction, the direction from a, to be and the direction from a to b is this angle.

04:53 We're going to call it alpha.

04:54 So that's the angle you're looking for the direction from a to b.

04:58 So we're looking for the distance a b and the direction from a to b.

05:05 One of the things that you're going to pick up is the fact that this whole so the distance is covered on a triangle, so it's going to be covered on a triangle.

05:24 So it's going to be covered on a triangle.

05:27 So that distance, i could say this is a, b relative to x, and this is a, b relative to y.

05:37 And then the distance we're looking for is a, b.

05:40 You know that's that's the distance you're pretty much looking for on the total distance is going to be a summation so the distance distance from a to b which i just called a b is a component of the three distances that you're looking for and then also same distance the displacement of the magnitude, the magnitude of ab, the magnitude of ab, which i'm calling ab magnitude is the square root.

06:28 So here we're using pythagoras of the components in the x direction squared plus the components in the y direction squared.

06:39 So that's what you're looking at in the particular problem.

06:42 So in terms of displacement, we do have the y direction and we also have the x direction.

07:03 So the displacement for 1 is d1, y hat, the displacement for 2.

07:14 These are displacements that you're looking at the displacements for the second one is remember it goes in the negative direction so this point here is going in the negative x direction and then this one is in the positive y direction so we're going to use those two to determine the displacement so we have negative d2 cosine of 20 cosine of 20 that's in the x direction or alternatively, we could say the displacement is d2 in the x direction plus d2 in the y direction.

08:02 So come back, we see that this is negative d2...

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