A delivery man travels 50m north, 40m east, 20m south and then takes elevator 50 m up in the building. What is his final displacement from the origin?
ans-50 *1.414=70.7
2007-07-14 18:57:05 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The question is about vectors; because there are measurements that possess both length (or magnitude) AND direction.
Your answer is either wrong or incomplete, as there is no mention of angle/direction.
His displacement (or change in position between the start and finish of the motion) is:
In the horizontal plane -- 50 m N + 40 m E + 20 m S means he ends that at a point 30 m N and 40 m E of his origin - a 3, 4, 5 triangle, so the horizontal component of his final displacement is 50 m at an angle of 52 degrees 8 minutes east of north.
In the vertical plane -- 50 m at the 52 degree 8 minutes angle across from the start, then 50 metres vertically gives a displacement (the hypotenuse of the triangle), as 70.7 (or 50root2) m at an angle of 45 degrees to the horizontal.
So your full answer should be 70.7 m bearing N 36degrees52minutes E, 45 degrees to horizontal.
2007-07-14 19:22:30 · answer #1 · answered by big_george 5 · 0⤊ 0⤋
This problem is three dimensional, meaning you need three parameters to desribe it. Let them be the distances north, east, and above the origin.
The man goes (50m - 20m) = 30m north
The man goes 40m east.
The man goes 50m up.
The distance is
d^2 = (30 ^2 + 40^2 + 50^2)m^2
d = [(5000)m^2]^(1/2)
d = 70.7m
2007-07-15 02:28:06 · answer #2 · answered by supastremph 6 · 0⤊ 0⤋
70.71
YOu've already said the answer, why asking ?
2007-07-15 02:05:56 · answer #3 · answered by Chris Chase 3 · 0⤊ 0⤋