In the figure, a radar station detects an airplane approaching directly from the east. At first observation, the airplane is at distance d1 = 320 m from the station and at angle θ1 = 31° above the horizon. The airplane is tracked through an angular change Δθ = 124° in the vertical east–west plane; its distance is then d2 = 800 m. Find the (a) magnitude and (b) direction of the airplane's displacement during this period. Give the direction as an angle relative to due west, with a positive angle being above the horizon and a negative angle being below the horizon.

The magnitude and the direction of the displacement is solved by the polar equation
800 ∠ 124° - 320 ∠ 31°
A capable calculator can solved this immediately.
Manually, the x and y components need to be solved using trigonometry. Then, by the Pythagorean theorem, the magnitude is solved. The direction is determined using the x and y components and the tangent function.

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aristocles aristocles · Answer 2 · 2019-10-05T22:44:21-04:00

Answer:

a) magnitude = 1014.2 m

b) angle = 170.16 degree

Explanation:

Initial position of the airplane is given as

\(d_1 = 320 m\) at an angle of 31 degree

so its position vector is given as

\(r_1 = 320 cos31\hat i + 320 sin31\hat j\)

\(r_1 = 274.3 \hat i + 164.8 \hat j\)

now for the final position we know that

\(d_2 = 800 m\) at an angle of (124+31) degree

\(d_2 = 800 m\) at angle 155 degree

final position vector is given as

\(r_2= 800 cos155\hat i + 800 sin155\hat j\)

\(r_2 = -725\hat i + 338.1 \hat j\)

now displacement is given as

\(d = r_2 - r_1\)

\(d = (-725 - 274.3)\hat i + (338.1 - 164.8)\)

\(d = -999.3\hat i + 173.3\hat j\)

magnitude of the displacement is given as

\(d = \sqrt{(-999.3)^2+ (173.3)^2}\)

\(d = 1014.2 m\)

direction is given as

\(tan\theta = \frac{173.3}{-999.3}\)

\(\theta = 170.2\) degree