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Original Question
taylor decides to search for hard evidence to support his "they never landed on the moon theory" so he gets into his cactus pow...Asked by jayven march
Taylor decides to search for hard evidence to support his "they never landed on the moon theory" so he gets into his cactus powered mobile and drives. he begins by driving 50.0km [N]. he then turns and drives 30.0k [N30E]. he comes to his futile search by driving 25.0 km [w] and then 40.0 km [sw]. if Taylors entire trip took 8.5 hours, find his average velocity.
his displacement that I got was 61 km and his average speed was 4.7m/s. if you needed to know
his displacement that I got was 61 km and his average speed was 4.7m/s. if you needed to know
Answers
Answered by
Reiny
I translated all to vectors:
50.0km [N] ---> 50(cos90,sin90) = (0,50)
30.0k [N30E] ---> 30(cos60,sin60) = (15,25.9808)
25.0 km [w] --> 25(cos180,sin180) = (-25, 0)
40.0 km [sw] --- 40(cos225,sin225) = (-28.28427, -28.28427)
resultant vector = (-38.28427, 47.69653)
magnitude = √((-38.28427)^2 + 47.69653^2) = 61.16 km
velocity = distance/time = 61.16 km/8.5 hrs = 7.195 km/h or appr 7.2 km/h
50.0km [N] ---> 50(cos90,sin90) = (0,50)
30.0k [N30E] ---> 30(cos60,sin60) = (15,25.9808)
25.0 km [w] --> 25(cos180,sin180) = (-25, 0)
40.0 km [sw] --- 40(cos225,sin225) = (-28.28427, -28.28427)
resultant vector = (-38.28427, 47.69653)
magnitude = √((-38.28427)^2 + 47.69653^2) = 61.16 km
velocity = distance/time = 61.16 km/8.5 hrs = 7.195 km/h or appr 7.2 km/h
Answered by
oobleck
the sum of his displacement vectors is
<0,50> + <30*0.5,30*0.866> + <-25,0> - <40*0.707 + 40*0.707> = <-38.28,47.7>
So your final displacement and speed have the right magnitude, but we are dealing with vectors here. The average velocity is
4.7 m/s (16.92 km/hr) in the direction N 38.75° W
<0,50> + <30*0.5,30*0.866> + <-25,0> - <40*0.707 + 40*0.707> = <-38.28,47.7>
So your final displacement and speed have the right magnitude, but we are dealing with vectors here. The average velocity is
4.7 m/s (16.92 km/hr) in the direction N 38.75° W
Answered by
oobleck
oops. I did not check your division to verify the speed.
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