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lesya692 [45]
3 years ago
7

A plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the grou

nd. What is the magnitude of the plane's total displacement? Answer in units of km.
Physics
1 answer:
Maslowich3 years ago
5 0

Answer:

6.1 km

Explanation:

Given that a plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the ground. What is the magnitude of the plane's total displacement? Answer in units of km

The magnitude of the total displacement D can be calculated by using cosine formula

Ø = 25 - 16 = 9 degree

D^2 = 4^2 + 10^2 - 2 × 4 × 10 × cos 9

D^2 = 16 + 100 - 80cos9

D^2 = 116 - 79.02

D = sqrt( 36.98)

D = 6.1 m

Therefore, the magnitude of the plane's total displacement is 6.1 km

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First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

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Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

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To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

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