Presentation is loading. Please wait.

Presentation is loading. Please wait.

The Long John Silver Solution (the easy  way)

Similar presentations


Presentation on theme: "The Long John Silver Solution (the easy  way)"— Presentation transcript:

1 The Long John Silver Solution (the easy  way)
B E C A D

2 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 left and down. Since you start at W(45, 30), then you finish at X(26.66, 16.66). Step #3: Need to go right and down to get from X to D. 1/4 of this is 3.33 right and 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).

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

4 Step #2: Move toward tree C, covering 1/3 the distance.
Step #1: Start at tree A and move toward tree B, covering only 1/2 the distance. G: the origin First, find the position of B relative to A. rBA = rBG + rGA = rBG - rAG = (60 i + 80j) - ( 30 i – 20 j) = 30 i + 100j (1/2) * rBA = 15 i + 50j rS1G = position at the end of step #1 = rAG + (15 i + 50 j) = 45 i j Step #2: Move toward tree C, covering 1/3 the distance. First, find the position of C relative to rS1G. rCS1 = rCG + rGS1 = rCG – rS1G = (-10 i j) - ( 45 i + 30 j) = -55 i – 40 j (1/3) * rCS1 = i j rS2G = position at the end of step #2 = rS1G + ( i j) = i j

5 Step #3: Move toward tree D, covering 1/4 the distance.
First, find the position of C relative to rS2G. rDS2 = rDG + rGS2 = rDG – rS2G = (40 i j) – ( i j) = i – j (1/4) * rDS2 = i j rS3G = position at the end of step #3 = rS2G + ( i j) = 30 i + 5 j Step #4: Move toward tree D, covering 1/5 the distance. First, find the position of E relative to rS3G. rES3 = rEG + rGS3 = rEG – rS3G = (-70 i + 60j) - ( 30 i + 5 j) = -100 i + 55 j (1/5) * rES3 = -20 i j rS2G = position at the end of step #1 = rS1G + (-20 i j) = 10 i j Dig here!!!

6 Problem #61 Solution ? = 106.26o + = 5 5 3 3 6 a q
These two x-components must cancel each other out sinq = 3/5  q = sin-1(3/5)=36.87o (notice that a equals the same value) ? ? = o 36.87 36.87


Download ppt "The Long John Silver Solution (the easy  way)"

Similar presentations


Ads by Google