Asked by Alex
An aircraft covers a distance of 400km; 60 east of south and then flew 500km, 45 east of north. Calculate the approximate resultant displacement
Answers
Answered by
Reiny
60 east of south ---> S60°E ----> 330° in standard math notation
45 east of north ---> N45°E ----> 45° in standard math notation
using vectors
R = 400(cos330°, sin330°) + 500(cos45°,sin45°)
= (346.41.., -200) + (353.5534,353.5534)
= (699.9636, 153.5534)
|R| = appr 716.6 km
angle of R
tanØ = 153.5534/699.9636
= .2193...
Ø = appr 12.37°
or N 77.6° E
or by cosine law, after making a sketch and calculating the angles,
R^2 = 400^2 + 500^2 - 2(400)(500)cos105°
= 513,527.618
R = √513,527.618 = 716.6 km , just like above
45 east of north ---> N45°E ----> 45° in standard math notation
using vectors
R = 400(cos330°, sin330°) + 500(cos45°,sin45°)
= (346.41.., -200) + (353.5534,353.5534)
= (699.9636, 153.5534)
|R| = appr 716.6 km
angle of R
tanØ = 153.5534/699.9636
= .2193...
Ø = appr 12.37°
or N 77.6° E
or by cosine law, after making a sketch and calculating the angles,
R^2 = 400^2 + 500^2 - 2(400)(500)cos105°
= 513,527.618
R = √513,527.618 = 716.6 km , just like above
Answered by
henry2,
All angles are measured CW from +y-axis.
Disp. = 400km[120o] + 500km[45o].
X = 400*sin120 + 500*sin45 = 700 km.
Y = 400*Cos120 + 500*Cos45 = 154 km.
Disp. = 700 + 154i = 717km[78o] CW.
Disp. = 400km[120o] + 500km[45o].
X = 400*sin120 + 500*sin45 = 700 km.
Y = 400*Cos120 + 500*Cos45 = 154 km.
Disp. = 700 + 154i = 717km[78o] CW.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.