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Hoochie [10]
4 years ago
11

A plane leaves the airport in Galisteo and flies 170 km at 68.0° east of north; then it changes direction to fly 230 km at 36.0°

south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?
Physics
1 answer:
luda_lava [24]4 years ago
4 0

Answer:

The direction will be 84.86^\circ and the distance 250.75km.

Explanation:

Let's say A is the displacement vector which represents the first 170km and B the one for the next 230km. Then the components of these vector will be:

A_x=170cos(68^{\circ})\\ A_y=170sin(68^\circ)\\\\B_x=230cos(-36^\circ)\\B_y=230sin(-36^\circ)

The vector which point from the origin to the final position of the plane will be R=A+B. We sum components on <em>x </em>and <em>y </em>independetly (vector property):

R_x=A_x+B_x=170cos(68^{\circ})+230cos(-36^\circ)=63.68km+186.07km=249.75km

R_y=A_y+B_y=170sin(68^\circ)+230sin(-36^\circ)=157.62km-135.19km=22.43km

If \theta is the direction of R then:

tan(\theta )=\frac{R_x}{R_y} ⇒ \theta = arctan(\frac{R_x}{R_y}) ⇒ \theta = 84.86^\circ.

The distance will be given by the magnitud of the vector R:

R=\sqrt{R_x^2 + R_y^2} ⇒ R=\sqrt{R_x^2 + R_y^2} = 250.75.

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