Respuesta :
Answer:
displacement is 481.3 m with 40.88° north east
Explanation:
given data
walk 1 = 300 m north = 300 j
walk 2 = 400 m northwest =400 cos(45) i + sin(45) j
walk 3 = 700 m east southeast = 700 cos(22.5) i- sin(22.5) j
to find out
displacement and direction
solution
we consider here direction x as east and direction y as north
so
displacement = distance 1 + distance 2 + distance 3
displacement = 300 j + 400 cos(45) i + sin(45) j + 700 cos(22.5) i- sin(22.5) j
we get here
displacement = 363.9 i + 315 j
so magnitude
displacement = [tex]\sqrt{363.9^{2}+315^{2} }[/tex]
displacement = 481.3 m
and angle will be = arctan(315/369.9)
angle = 0.713494059 rad
angle is 40.88 degree
so displacement is 481.3 m with 40.88° north east
Answer:
Magnitude of displacement = 481.24 m
Direction = 40.88 degrees north east.
Explanation:
Let east is along real axis and north is along imaginary axis. So,
First walk = d1 = J300 m
Second walk = d2 = -400Cos(45) + J400Sin(45) = (-282.84 + J282.84) m
Third walk = d3 = 700Cos(22.5) – J700Sin(22.5) = (646.71 – J267.88) m
Total displacement = d = d1 + d2 + d2 = (363.86 + J314.96)m
Magnitude = √((363.86)^2+(314.96)^2) = 481.24 m
Direction = arctan(314.96/363.86) = 40.88 degrees north east.