Use either trigonometry or components to find displacement

AI Thread Summary
To find Bob's displacement returning to his starting point, the problem requires calculating the net x- and y-components of his movements. Bob walks 160 m south, jogs 630 m southwest, and then walks 160 m at 32 degrees east of north. The correct approach involves summing the individual components of each vector and applying the Pythagorean theorem to determine the straight-line distance. The direction of the displacement can be found using trigonometric functions based on the resultant components. This method ensures an accurate calculation of both distance and direction for Bob's return.
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Homework Statement


Bob walks 160 m south, then jogs 630 m southwest, then walks 160 m in a direction 32degrees east of north.

a) Use either trigonometry or components to find the displacement that will return Bob to his starting point by the most direct route. Give your answer as a distance.

b)Find the direction of the displacement that will return Bob to his starting point by the most direct route.


Homework Equations


Not sure what to use.


The Attempt at a Solution



I'm not even sure where to begin, even some hints to tell me how to start would be very useful!

a)
60 + 630(cos45+sin45) + 160(cos32,sin32)
60 + 630(0.71 + 0.71) + 160(0.848 + 0.5299)
60 + 447.3 + 447.3 + 135.688 + 84.787
642.98 + 533.087

642.98^2 + 533.087^2
413423.28 + 284181.83
697605
take root = 835.22
 
Last edited:
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Draw a picture of the displacement vectors. Bob can only move in two dimensions (a plane), so determine the x- and y-components of each vector. Now add the individual components together and you get Bob's net displacement from his initial position. It appears that you did this but I can't tell what the x- and y-components are from your notation. Also I see a 60 as the first time in your work, but not in the problem statement; should it be 160?

The most direct route is straight line, so for part a) apply the Pythagorean theorem to your net x- and y-components and take the square root to get the distance.
 
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