The minute hand on a watch is 2.0 cm in length. What is the displacement vector of the tip of the minute hand in each case? Use a coordinate system in which the y-axis points toward the 12 on the watch face. a. From 8:00 to 8:20 a.m.

Video transcript

Hey, everyone in this problem, we have a diver that has a watch that indicates his diving time is 20 seconds. It's a mechanical watch and the second hand is three centimeters long. We're gonna use a coordinate system in which the Y axis is aligned with the 62nd marker on the watch. We're asked to find the displacement vector of the second hand tip between zero and 20 seconds. And as we're given this little diagram, we have the watch, you can see that it started at 12 or 60 seconds and it's gone 20 seconds to the four second hand has and we have four answer choices all in centimeters. Option A 4.2 I hat minus 3.5 J hat. Option B 2.6 I hat minus 4.5 J hat. Option C negative 3.2 I hat plus 2.5 J hat and option D negative 1.3 I hat minus 0.5 J hat. So let's start by drawing out this Cartesian coordinate system that they want us to use. So we're gonna use this just as regular coordinate system. We have our X and Y axis. And the second hand, if we're at zero seconds and what's that gonna look like? Well, that's just gonna be pointing straight upwards at that 12 or at the 60 seconds. And we're gonna call that D zero. And so that is the second hand at times zero. And another hand is gonna travel all the way around until the four, maybe 20 seconds. And so we're gonna call that D 20. So we have these two vectors and we wanna find the displacement vector of the tip of this second hit and the displacement of that tip. Well, we're gonna draw a vector that connects the tip of D zero to the tip of D 20. And that is our displacement vector D that we're interested in. All right. So that line looks a little pink. It should be perfectly straight there. OK? So we have our displacement vector D. Now the angle between our vector D zero and D 20 we're gonna call that alpha. And we're gonna need that in order to calculate this. And you'll see why in just a minute. So how can we calculate the displacement vector? Right? Well, we can do a little bit of vector math here. What we can see is that if we were to add the negative of D zero. OK. So the negative of D zero, it would be the same vector, but it would be pointing in the opposite direction. So we add that vector to D 20 then we add tip to tail and we get D like we want. OK. So what that tells us is that D is going to be negative D zero. UD 20. OK. So we're taking D 20 subtracting D zero and we're gonna get that displacement vector that we want. D OK. So what's D zero? Well, D zero is easy because we know it's pointing straight up along that Y axis. OK. So the horizontal component is going to be zero. So we have D zero has a zero, horizontal component that means zero. I ha you recall that I hack corresponds to the horizontal or X direction J hat corresponds to the vertical or Y direction. And then in the Y direction will we know that the second hand is three centimeters long. So we're gonna have three in the vertical component and D zero has a length of three. So we get three J hat and this is in centimeters. So we have D zero. But in order to find D, we need to know D 20. Now, we know the length of D 20. This is that the second hand, we know it's three centimeters long, but we don't know the two components. OK. So let's go ahead and just draw out our vector D 20 on its own and see what we can find. So we have a Rector D 20 and we know that it is three centimeters long and it's gonna have some X component, which we're gonna call DX and it's also gonna have some Y component, which we're gonna call Dy. We're gonna say that the angle between the horizontal and D 20 is theta. Now, in order to calculate DX and Dy, we need to find the, OK. Let's think about what we have here. We can see that the total angle alpha between D zero and D 20 we can calculate that because we know the time taken to get to D 20 we know that that covers 20 seconds, ok? So let's go ahead and calculate alpha first. So alpha, so we know that in an entire rotation, we have 360 degrees. And in this watch, we have 60 seconds. So in 360 degrees, we cover 60 seconds. So we're gonna take 360 degrees divided by 60 seconds and then we're gonna multiply by 20 seconds. The amount of time that our hand actually moves, the unit of seconds will divide it. And we get that alpha is going to be equal to 120 degrees. And this makes sense. 20 seconds is gonna be one third of the rotation, one third of a minute. So it goes one third around. So 120 is one third of the 360 degrees that make up that entire circle. Ok? So we have our alpha value. What we can see now is that theta is going to be related to alpha. Alpha is the entire angle from D zero down to D 20. And theta is just the angle from the horizontal down to D 20. So theta is gonna be equal to that 120 degrees minus 90 degrees and that 90 degrees comes from the entire first quad. Let me draw theta. I'm gonna draw it in blue and I'm gonna draw it on our first diagram. So we can really see that relationship. OK. So the is just from the horizontal. And so if we take our alpha value of 120 degrees, we subtract 90 degrees, we are gonna get our theta value which is going to be equal to 30 degrees. Yeah. All right. So a lot going on. But so far just triangle math, we're just working at these angles, these side links. No problem. Now we know theta, we know the hypotenuse we can use our trigonometric ratios because this is a right angle triangle to find DX and Dy. So let's first think about sign. OK. If we think about sign of the angle, this is gonna relate the opposite side to the hypotony S data is equal to the opposite divided by the hypo. In this case, we have sign of 30 degrees, the opposite side is going to be Dy. And we know that this is going to be negative because it's going the vector would point downwards OK. In that negative direction, the hypotenuse is three centimeters. And so if we multiply three centimeters on both sides and simplify, we get that dy is going to be negative 1.5 centimeters. Let's do the same for cosine. OK. So that we can get the X cosine is gonna be the adjacent side divided by the hippot. So cosine of 30 degrees is going to be equal to the X divided by three centimeters. Again, multiplying both sides by three centimeters. We get the DX that's gonna be approximately equal to 2.6. All right. So we have now RDY and DX values four D 20. Now remember we're trying to figure out what the displacement vector D is, we know D zero, but we need ad 20 in order to calculate D. So now we have the components of D 20 we can solve for D. So let's get back to that. So we now can write that D 20 is going to be equal to 2.6 I hat minus 1.5 J hat centimeters. Recall that we had D as negative D zero last 20. OK. This is going to be negative zero I ha plus three J hat centimeters of course, 2.6 I hat minus 1.5 J hat centimeters. When we're adding vectors like this, we wanna add the corresponding components and, and we get what? Well, the X component of the first vector is just zero, the X component of the second vector 2.6 I, so we have D is going to be 2.6 I. Then we're gonna do the Y components in the first vector. We have negative three. They don't forget to distribute this negative to both terms. In the second, we have negative 1.5 adding those together, we get negative 4.5. And so our vector D, the displacement vector of that second hand that we were looking for is gonna be 2.6 I hat minus 4.5 J hat centimeters. If we compare this to our answer choices, we can see that this corresponds with answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.