Respuesta :
Velocity of plane is given as
[tex]v_1 = 2.50 * 10^2 km/h[/tex] towards north
[tex]v_2 = 75 km/h[/tex] towards south east
we can write the two velocity in vector form
[tex]v_1 = 2.50 * 10^2 \hat j[/tex]
[tex]v_2 = 75 cos45 \hat i - 75 sin45 \hat j[/tex]
now the net velocity will be given as
[tex]v_{net} = v_1 + v_2[/tex]
[tex]v_{net} = 250\hat j + 53.03 \hat i - 53.03 \hat j[/tex]
[tex]v_{net} = 53.03\hat i + 196.97 \hat j[/tex]
so the resultant velocity will be given as
[tex]v_{net} = \sqrt{53.03^2 + 196.97^2}[/tex]
[tex]v_{net} = 203.98 m/s[/tex]
and the angle of velocity will be
[tex]\theta = tan^{-1}\frac{v_y}{v_x} [/tex]
[tex]\theta = tan^{-1}\frac{196.97}{53.03} = 74.9 degree[/tex]
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, [tex]v_{plane}[/tex] = 2.50 × 10² km/h
Direction of the plane = North
Velocity of the wind, [tex]v_{wind}[/tex] = 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/s
- The 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;
- [tex]\mathbf{v_{plane}}[/tex] = 2.5×10²·j
- [tex]\mathbf{v_{wind}}[/tex] = 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 ≈ [tex]arctan \left(\dfrac{196.97}{53.03} \right) \approx 74.9^{\circ}[/tex]
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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