Physics, 09.07.2019 04:00 isaiahdunn

Anovice pilot sets a plane’s controls, thinking the plane will fly at 2.50 × 102 km/h to the north. if the wind blows at 75 km/h toward the southeast, what is the plane’s resultant velocity? use graphical techniques.

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Answer from: jonjosef2012

The magnitude and direction of the resultant velocity of the airplane

obtained graphically are 201.43 km/hr in the direction North 15.41° West,

The given parameters are;

Velocity of the plane, $v_{plane}$ = 2.50 × 10² km/h

Direction of the plane = North

Velocity of the wind, $v_{wind}$ = 75 km/h

Direction of the wind = Southeast

Required:

Resultant velocity of the plane

Solution:

Using graphical techniques, we have;

The scale of the vector drawing is 1 mm = 1 m/sThe line representing the velocity of the plane is a vertical line 250 mm long.The line representing the magnitude of the velocity of the wind is a line 75 mm long, rotated 45° clockwise from the from the positive x-axis.The end of the wind vector line is joined by a line drawn from beginning of the plane vector line, to represent the resultant velocity of plane.

The direction of the plane relative to the x-axis is approximately 74.59°,

which is (90° - 74.59° ≈ 15.41°) North 15.41° West

Please find attached the drawing of the resultant velocity of the plane.

By calculation, we have;

The component form of the given velocities are;

$\mathbf{v_{plane}}$ = 2.5×10²·j $\mathbf{v_{wind}}$ = 75 × cos(45°)·i - 75 × sin(45°)·j

The resultant velocity is given by the sum of the two velocities as follows;

v = (75 × cos(45°) + 0)·i + (2.5×10² - 75 × sin(45°))·j ≈ 53.03·i + 196.97·j

The magnitude of the velocity, v = √(53.03² + 196.97²) ≈ 203.98

The plane resultant velocity, v ≈ 203.98 km/h

The direction of the resultant velocity, θ ≈ $arctan \left(\dfrac{196.97}{53.03} \right) \approx 74.9^{\circ}$

The magnitude and direction of the resultant velocity of the airplane are;

Magnitude of resultant velocity, v ≈ 203.98 km/hr, direction, θ ≈ 74.9°, which is North 15.1° West

Therefore, there is a slight error in the values obtained graphically.

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Answer from: annie1799

Velocity of plane is given as

v_1 = 2.50 * 10^2 km/h towards north

v_2 = 75 km/h towards south east

we can write the two velocity in vector form

$v_1 = 2.50 * 10^2 \hat j$

$v_2 = 75 cos45 \hat i - 75 sin45 \hat j$

now the net velocity will be given as

$v_{net} = v_1 + v_2$

$v_{net} = 250\hat j + 53.03 \hat i - 53.03 \hat j$

$v_{net} = 53.03\hat i + 196.97 \hat j$

so the resultant velocity will be given as

$v_{net} = \sqrt{53.03^2 + 196.97^2}$

$v_{net} = 203.98 m/s$

and the angle of velocity will be

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{196.97}{53.03} = 74.9 degree$

Anovice pilot sets a plane’s controls, thinking the plane will fly at 2.50 × 102 km/h to the north.

Answer from: Quest

concave lenses are used here. they spread the light out before it reaches the convex lens in the eye, therefore letting the image focus directly on the retina. hyperopia is known to you probably as farsightedness.

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

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Anovice pilot sets a plane’s controls, thinking the plane will fly at 2.50 × 102 km/h to the north....