1. AP Physics Unit 2 - Vectors

2. Day #6 Relative Velocity and i/j/k notation

3. We’re here on your schedule

5. #59 The Long John Silver Solution (the easy way) A D B C E

6. Step #1: Need to go 30 right and 100 up to get from A to B. 1/2 of this is 15 right and 50 up. Since you start at A(30, -20), then you finish at W(45, 30). Step #2: Need to go 55 left and 40 down to get from W to C. 1/3 of this is 18.33 left and 13.33 down. Since you start at W(45, 30), then you finish at X(26.66, 16.66). Step #3: Need to go 13.33 right and 46.66 down to get from X to D. 1/4 of this is 3.33 right and 11.66 down. Since you start at X(26.66, 16.66), then you finish at Y(30, 5). Step #4: Need to go 100 left and 55 up to get from Y to E. 1/5 of this 20 left and 11 up. Since you start at Y(30, 5), then you finish at Z(10, 16).

7. The Long John Silver Solution (the physics way) A D B C E

8. Step #1: Start at tree A and move toward tree B, covering only 1/2 the distance. First, find the position of B relative to A. r BA = r BG + r GA = r BG - r AG = (60 i + 80j) - ( 30 i – 20 j) = 30 i + 100j (1/2) * r BA = 15 i + 50j r S1G = position at the end of step #1 = r AG + (15 i + 50 j) = 45 i + 30 j Step #2: Move toward tree C, covering 1/3 the distance. First, find the position of C relative to r S1G. r CS1 = r CG + r GS1 = r CG – r S1G = (-10 i - 10 j) - ( 45 i + 30 j) = -55 i – 40 j (1/3) * r CS1 = -18.33333 i - 13.33333 j r S2G = position at the end of step #2 = r S1G + (-18.33333 i - 13.33333 j) = 26.6666 i + 16.66666 j G: the origin

9. Step #3: Move toward tree D, covering 1/4 the distance. Step #4: Move toward tree D, covering 1/5 the distance. First, find the position of E relative to r S3G. r ES3 = r EG + r GS3 = r EG – r S3G = (-70 i + 60j) - ( 30 i + 5 j) = -100 i + 55 j (1/5) * r ES3 = -20 i + 11 j r S2G = position at the end of step #1 = r S1G + (-20 i + 11 j) = 10 i + 16 j First, find the position of C relative to r S2G. r DS2 = r DG + r GS2 = r DG – r S2G = (40 i - 30 j) – (26.6666 i + 16.66666 j) = 13.33333 i – 46.66666 j (1/4) * r DS2 = 3.33333i - 11.66666 j r S3G = position at the end of step #3 = r S2G + (3.33333i - 11.66666 j) = 30 i + 5 j Dig here!!!

10. Problem #61 Solution     +  = sin  = 3/5   = sin -1 (3/5)=36.87 o (notice that  and  equals the same value)       These two x-components must cancel each other out

11. Read & learn the rest of Chapter 3. Solve problems 3, 10, 23, 28, 31, and 33 on pp 65-67. TONIGHTS HW

12. X: 50 +32cos40 – 16cos30 = 60.675 km Y: 0 +32sin40 – 16sin30 = 12.569 km  r = (60.675i + 12.569j) km  r| 61.96 km Since  r 61.96 km [E 11.7 o N], the car must travel _____________________ to get back. 61.96 km [W 11.7 o S]

13. 40 o 30 o 11.7 o

18. xx yy zz