Bubba decides to go fishing one day. He gets into his boat and starts out from t

Question

Bubba decides to go fishing one day. He gets into his boat and starts out from the boat ramp to find the perfect fishing spot in the lake. First, he goes 121 meters in a direction of 20 degree west of north. north. Then, he goes 104 meters in a direction of 33.0 degree west of meters in a direction of south of west. Finally, he turns and heads 135 meters in a direction of 41.8 degree north of east. He then realizes that he forgot his fishing pole. What direction and distance must he go in order to return directly to the boat ramp? If he were moving at 1.55 m/s the entire lime, how long did it take him to reach the spot where he realized that he had forgotten his fishing pole? If he heads back to the boat ramp on the most direct path at the same speed of 1.55 m/s, how long does it take him to return to the boat ramp?

Explanation / Answer

Q6.

let east be along +ve x axis and north be along +ve y axis.

let i and j are unit vectors along +ve x and +ve y axis respectively.

then his total displacement

=121*(-sin(20) i + cos(20) j) + 104*(-cos(33) i -sin(33) j) + 211*(cos(20.5) i - sin(20.5) j)+135*(cos(41.8) i+ sin(41.8) j)

=169.67 i + 73.148 j

part a:

in order to return directly to boat ramp, his displacement should be negative of his current position vector

so displacement=-168.67 i -73.148 j

so distance to be travelled=sqrt(169.67^2+73.148^2)=184.77 m

as it is third quadrant it makes angle theta south of west.

theta=arctan(73.148/168.67)=23.445 degrees

so he has to travel 184.77 m ,along 23.445 degrees south of west.

part b:

time taken=(121/1.55)+(104/1.55)+(211/1.55)+(135/1.55)=368.39 seconds

part c:

time taken in most direct path=184.77/1.55=119.21 seconds

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