Exercise 2 10 points a particle is moving due east with a...

Name: exercise 2 10 points a particle is moving due east with a speed of vi 8 ms then it turns heading northeast with a speed of vf 20 ms note northeast is 45 a using a scale 1 cm 2 ms sketch th 27832
Uploaded: 2021-06-29T15:27:00-08:00
Duration: 8 min 12 s
Channel: Thomas Thompson
Description: exercise 2 10 points a particle is moving due east with a speed of vi 8 ms then it turns heading northeast with a speed of vf 20 ms note northeast is 45 a using a scale 1 cm 2 ms sketch th 27832

Exercise 2] (10 points) A particle is moving due East with a speed of $v_i = 8 m/s$, then it turns heading Northeast with a speed of $v_f = 20 m/s$. (Note: Northeast is 45°) (a) Using a scale 1 cm = 2 m/s sketch these velocity vectors. (b) On the same figure, sketch $-v_i$ attached to the head of $v_f$. (c) On the same figure, sketch a vector from the tail of $v_f$ to the head of $-v_i$ and label it $\Delta v$. State the angle associated with $\Delta v$. (d) If the turn takes 2 seconds, what is the acceleration of this particle? (e) Sketch the associated acceleration vector (use 1 cm = 2 m/s² as your scale) on the same figure above. State the angle associated with the vector a. (f) If the mass of the particle is m = 0.5 kg, what is the net force acting during the turn? Find its magnitude and direction.

Transcript

00:01 This problem deals with displacement and its relationship to speed and to velocity.

00:07 We're going to deal with vectors which have magnitude and a direction and scalers like speed, which only have a magnitude.

00:15 I've written down the path that this motorist takes and i've put it onto the x, y axis or a grid here.

00:25 The main equation we're going to use is that velocity or speed, depending on whether it's a vector or scalar is change in displacement over change in time.

00:35 Now when i write down this information i can see that my units don't match and so immediately i'm going to write down time in seconds so i just multiply it by 60 seconds per minute and so i get 180 seconds and then down here i get 120 seconds and then going northwest i get 60 seconds.

00:56 Then i can use this equation right here and i can figure out what the displacement is in each case.

01:05 So over here i have 180 times 2, so i have 3 ,600 meters or 3 .6 kilometers.

01:13 Then down here, 120 seconds at 25 meters per second, so i have 3 ,000 meters.

01:20 And then going northwest, my time is 60 seconds.

01:27 And so i wrote this in the wrong place.

01:32 And that means that my displacement is 1 ,800 meters.

01:38 Now i can figure out what i need to do is find out what this displacement is right here.

01:45 I'm going to call that r sub t, the total displacement, the final displacement.

01:51 And i'll do that on the next screen.

01:54 So my end point here is going to be at the end of that last vector.

02:01 Now this point here is minus 3 ,600 from my first leg and minus 3 ,000 in the x direction from the westward leg.

02:13 I know that this final leg here is northwest, which means this angle is 45 degrees.

02:19 Now, if it has a distance of 1 ,800 meters, then that means that each side is 1 ,800 divided by the square root of 2, just basic trigonometry there.

02:30 And so each side is going to be 1272 .8 meters.

02:36 Now, if i look at the delta y, it's going back in the positive direction.

02:41 This leg here is going in the positive direction.

02:44 So i have plus 1272 .8 meters.

02:49 But then down here for x, this leg is going in the negative x direction.

02:55 So i have minus 1272 .8 meters.

02:58 And so therefore, therefore my displacement in the y direction ends up being a total, if i add those two together, of minus 23 -27 .2 meters.

03:17 And then in the x direction, i have minus 42 -72 .8.

03:23 So those are the final coordinates of this point right here.

03:28 Now if i want to find this, that final or total delta r, what i need to do is a little bit of pythagorean theorem, and i'll do that on the next slide.

03:43 So if i draw that delta rt vector, then i have a y displacement and i have an x displacement.

03:52 So my delta rt, if i use pythagorean theorem, because this is a right angle, is going to be the square root of delta x square...

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