Answer :
Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North
Explanation:
In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).
Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC
AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)
Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees
Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North
Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North
The Total displacement of the airplane is ; ≈ 172.19 km 34.2° east of North
The double displacement of the airplane forms a right angled triangle assuming the plane takes of at a point A
where ;
AC = total displacement
AB = fist displacement ( 75 km )
BC = second displacement ( 155 km ).
First step : calculate the value of the total displacement
∴ Total displacement = AC² = AB² + BC²
= 75² + 155² = 29650
hence AC = [tex]\sqrt{29650}[/tex] = 172.19 km
Next step ; determine the angle of the total displacement
The angle at point A = [tex]Tan^{-1} (155/75)[/tex] = 64.2°
therefore the direction of the total displacement will be
∠ AC = ( 64.2 ° - 30° ) = 34.2° east north
Hence we can conclude that the total displacement of the airplane after two ( 2 ) displacements is = 172.19 km 34.2° east of north.
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