Question

An airplane takes off and heads for a destination that is 673 km away in a direction 72.0° North of West. In order to avoid a storms, the airplane initially flies 128 km at 7.0° East of North before changing direction and heading straight to the final destination. How far and in which direction should the airplane fly during the second leg of its trip in order to reach its final destination safely?

Answer 1

let East be +x axis

Rx = -R*cos(72)

= -673*cos(72)

= -208 m

Ry = R*sin(72)

= 673*sin(72)

= 640 km

A = 128 km

Ax = 128*sin(7)

= 15.6 km

Ay = 128*cos(7)

= 127 km

let B = Bxi + By j

Rx = Ax + Bx

==> Bx = Rx - Ax

= -208 - 15.6

= -223.6 km

By = Ry - Ay

= 640 - 127

= 513 km

so, B = sqrt(Bx^2 + By^2)

= sqrt(223.6^2 + 513^2)

= 560 km <<<<<<<<<----------------------Answer

direction : theta = tan^-1(By/Bx)

= tan^-1(513/223.6)

= 66.4 degrees North of West <<<<<<<<<----------------------Answer