1. 1 Relative Velocities in 1 D Schmedrick and his dog, Rover, are goofing around on a train. Schmed can throw a fast ball at 23 m/s. Rover can run at 9 m/s. The train goes 15 m/s. continued on next slide Question 1: If Rover is sitting beside the tracks with a radar gun as the train goes by, and Schmedrick is on the train throwing a fastball in the direction of the train, how fast does Rover clock the ball? v BT = velocity of the ball with respect to the train = 23 m/s v TG = velocity of the train with respect to the ground = 15 m/s v BG = velocity of the ball with respect to ground = 38 m/s This is a simple example, but in general, to get the answer we add vectors: v BG = v BT + v TG (In this case we can simply add magnitudes since the vectors are parallel.)

2. 2 Relative Velocities in 1 D (cont.) Velocities are not absolute; they depend on the motion of the person who is doing the measuring. Write a vector sum so that the inner subscripts match. The outer subscripts give the subscripts for the resultant. This trick works even when vectors don’t line up. Vector diagrams help (especially when we move to 2-D). v BG = v BT + v TG v BT = 23 m/s v TG = 15 m/s v BG = 38 m/s continued on next slide

3. 3 Question 2: Let’s choose the positive direction to be to the right. If Schmedrick is standing still on the ground and Rover is running to the right, then the velocity of Rover with respect to Schmedrick = v RS = +9 m/s. From Rover’s perspective, though, he is the one who is still and Schmedrick (and the rest of the landscape) is moving to the left at 9 m/s. This means the velocity of Schmedrick with respect to Rover = v SR = -9 m/s. Therefore, v RS = - v SR The moral of the story is that you get the opposite of a vector if you reverse the subscripts. Relative Velocities in 1 D (cont.) continued on next slide v SR v RS

4. 4 Relative Velocities in 1 D (cont.) Question 3: If Rover is chasing the train as Schmed goes by throwing a fastball, at what speed does Rover clock the ball now? v BT = 23 m/s v TG = 15 m/s v BG = 29 m/s Note, because Rover is chasing the train, he will measure a slower speed. (In fact, if Rover could run at 38 m/s he’d say the fastball is at rest.) This time we need the velocity of the ball with respect to Rover: v BR = v BT + v TG + v GR = v BT + v TG - v RG = 23 + 15 - 9 = 29 m/s. Note how the inner subscripts match up again and the outer most give the subscripts of the resultant. Also, we make use of the fact that v GR = - v RG. v RG = 9 m/s

5. 5 River Crossing Current 0.3 m/s campsite boat You’re directly across a 20 m wide river from your buddies’ campsite. Your only means of crossing is your trusty rowboat, which you can row at 0.5 m/s in still water. If you “aim” your boat directly at the camp, you’ll end up to the right of it because of the current. At what angle should you row in order to trying to land right at the campsite, and how long will it take you to get there? river continued on next slide

6. 6 River Crossing (cont.) Current 0.3 m/s campsite boat river 0.3 m/s 0.5 m/s Because of the current, your boat points in the direction of red but moves in the direction of green. The Pythagorean theorem tells us that green’s magnitude is 0.4 m/s. This is the speed you’re moving with respect to the campsite. Thus, t = d / v = (20 m) / (0.4 m/s) = 50 s.  = tan -1 (0.3 / 0.4)  36.9 .  0.4 m/s continued on next slide

7. 7 River Crossing: Relative Velocities Current 0.3 m/s campsite river 0.3 m/s 0.5 m/s  0.4 m/s The red vector is the velocity of the boat with respect to the water, v BW, which is what your speedometer would read. Blue is the velocity of the water w/ resp. to the camp, v WC. Green is the velocity of the boat with respect to the camp, v BC. The only thing that could vary in our problem was . It had to be determined so that red + blue gave a vector pointing directly across the river, which is the way you wanted to go. continued on next slide

8. 8 River Crossing: Relative Velocities (cont.) v WC v BW  v BC v BW = vel. of boat w/ respect to water v WC = vel. of water w/ respect to camp v BC = vel. of boat w/ respect to camp v BW + v WC = v BC Look how they add up: The inner subscripts match; the out ones give subscripts of the resultant. This technique works in 1, 2, or 3 dimensions w/ any number or vectors.

9. 9 Law of Sines The river problem involved a right triangle. If it hadn’t we would have had to use either component techniques or the two laws you’ll also do in trig class: Law of Sines & Law of Cosines. Law of Sines: sin A sin B sin C a b c = = Side a is opposite angle A, b is opposite B, and c is opposite C. AB C c b a

10. 10 Law of Cosines Law of Cosines: a 2 = b 2 + c 2 - 2 b c cosA This side is always opposite this angle. These two sides are repeated. It doesn’t matter which side is called a, b, and c, so long as the two rules above are followed. This law is like the Pythagorean theorem with a built in correction term of -2 b c cos A. This term allows us to work with non-right triangles. Note if A = 90 , this term drops out (cos 90  = 0), and we have the normal Pythagorean theorem. A B C c b a

11. 11 v WA = vel. of Wonder Woman w/ resp. to the air v AG = vel. of the air w/ resp. to the ground (and Aqua Man) v WG = vel. of Wonder Woman w/ resp. to the ground (Aqua Man) Wonder Woman Jet Problem Suppose Wonder Woman is flying her invisible jet. Her onboard controls display a velocity of 304 mph 10  E of N. A wind blows at 195 mph in the direction of 32  N of E. What is her velocity with respect to Aqua Man, who is resting poolside down on the ground? We know the first two vectors; we need to find the third. First we’ll find it using the laws of sines & cosines, then we’ll check the result using components. Either way, we need to make a vector diagram. continued on next slide

12. 12 The 80  angle at the lower right is the complement of the 10  angle. The two 80  angles are alternate interior. The 100  angle is the supplement of the 80  angle. Now we know the angle between red and blue is 132 . Wonder Woman Jet Problem (cont.) continued on next slide 10  32  v WA v AG v WG v WA + v AG = v WG 80  195 mph 304 mph v WG 80  32  100 

13. 13 Wonder Woman Problem: Component Method 32  v WA = 304 mph v AG = 195 mph 10  This time we’ll add vectors via components as we’ve done before. Note that because of the angles given here, we use cosine for the vertical comp. of red but sine for the vertical comp. of blue. All units are mph. 304 195 103.3343 165.3694 52.789 299.3816 continued on next slide

14. 14 Wonder Woman: Component Method (cont.) 304 195 103.3343 165.3694 52.789 299.3816 103.3343 52.789 165.3694 299.3816 402.7159 mph 218.1584 mph 458.0100 mph Combine vertical & horiz. comps. separately and use Pythag. theorem.  = tan -1 (218.1584 / 402.7159) = 28.4452 .  is measured from the vertical, which is why it’s 10  more than . 

15. 15 Comparison of Methods We ended up with same result for Wonder Woman doing it in two different ways. Each way requires some work. You will only want to use the laws of sines & cosines if: the vectors form a triangle. you’re dealing with exactly 3 vectors. (If you’re adding 3 vectors, the resultant makes a total of 4, and this method would require using 2 separate triangles.) Regardless of the method, draw a vector diagram! To determine which two vectors add to the third, use the subscript trick.

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