An aircraft covers a distance of 400km; 60 east of south and then flew 500km, 45 east of north. Calculate the approximate resultant displacement
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
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.
To find the resultant displacement, we need to find the sum of the two displacements. Each displacement can be represented as a vector.
The first displacement is 400 km, 60° east of south. To represent this as a vector, we can break it down into its north-south (vertical) and east-west (horizontal) components.
The north-south component can be found using trigonometry: sinθ = Opposite/Hypotenuse. In this case, the hypotenuse is the magnitude of the displacement (400 km) and the opposite side is the north-south component (let's call it y). So, sin(60) = y/400.
Solving for y, we get y = 400*sin(60) = 346.41 km.
The east-west component can also be found using trigonometry: cosθ = Adjacent/Hypotenuse. In this case, the adjacent side is the east-west component (let's call it x). So, cos(60) = x/400.
Solving for x, we get x = 400*cos(60) = 200 km.
Therefore, the first displacement can be represented as a vector: (x, y) = (200 km, -346.41 km). Note that the negative sign for the y-component indicates a southward direction.
The second displacement is 500 km, 45° east of north. Again, we can break it down into its north-south and east-west components.
Using the same trigonometric approach, we find the north-south component (let's call it y') is y' = 500*sin(45) = 353.55 km.
And the east-west component (let's call it x') is x' = 500*cos(45) = 353.55 km.
Therefore, the second displacement can be represented as a vector: (x', y') = (353.55 km, 353.55 km).
Now, to find the resultant displacement, we can simply add the corresponding components of the two vectors:
Resultant x-component: x + x' = 200 km + 353.55 km = 553.55 km (to the east)
Resultant y-component: y + y' = -346.41 km + 353.55 km = 7.14 km (to the north)
So, the approximate resultant displacement is a vector: (553.55 km, 7.14 km).