A plane is flying at 25 degrees north of west at 190km/ h. Suddenly there is a wind from 15degrees north of east at 45 km/ hr. what is the planes new velocity with respect to ground in standard position
V = 190km/h @ 155o + 45km/h @ 15o.
X = 190*cos155 + 45*cos15=-128.73 km/h.
Y = 190*sin155 + 45*sin15=91.94 km/h.
tanAr = Y/X = 91.94/-128.73 = 0.71424
Ar = -35.53o = Reference angle.
A = -35.54 + 180 = 144.5o
V = X/cosA = -128.73/cos144.5=158.1 km/h
@ 144.5o, CCW.
Note: One student said that the book's
answer is 227 km/h @ 162.4o. Based on
the INFO given, that would be impossible
because:
1. The direction of the wind reduces
the speed of the plane.
2. The resultant direction would have to be less than 155o.
To find the plane's new velocity with respect to the ground, we need to use vector addition. Let's break down the given information into components.
The plane's velocity is given as 190 km/hr at 25 degrees North of West. To convert this into components, we can split it into its North and West components.
North component: 190 km/hr * sin(25°)
West component: 190 km/hr * cos(25°) * -1 (since it is West of North)
The wind's velocity is given as 45 km/hr at 15 degrees North of East. To convert this into components, we can split it into its North and East components.
North component: 45 km/hr * sin(15°)
East component: 45 km/hr * cos(15°)
Next, we can add the North and East components separately. Adding the North component of the plane's velocity to the North component of the wind's velocity, and adding the West component of the plane's velocity to the East component of the wind's velocity.
New North component: (190 km/hr * sin(25°)) + (45 km/hr * sin(15°))
New East component: (45 km/hr * cos(15°)) - (190 km/hr * cos(25°))
To find the magnitude and direction of the new velocity, we can use the Pythagorean theorem and inverse tangent function.
Magnitude of the new velocity: sqrt((New North component)^2 + (New East component)^2)
Direction of the new velocity (with respect to the positive x-axis): arctan(New North component / New East component)
Calculating the above expressions will give us the plane's new velocity with respect to the ground in standard position.
To find the plane's new velocity with respect to the ground, we need to use vector addition. We'll break down the given velocities into their North and West components, and then add them up.
The plane's velocity is given as 190 km/h at 25 degrees north of west. To find the North and West components, we'll use trigonometry.
The North component of the plane's velocity is given by:
North component = Velocity * sin(Angle)
North component = 190 km/h * sin(25°)
Using a calculator, we find:
North component = 190 km/h * 0.4226
North component ≈ 80.39 km/h (rounded to two decimal places)
The West component of the plane's velocity is given by:
West component = Velocity * cos(Angle)
West component = 190 km/h * cos(25°)
Using a calculator, we find:
West component = 190 km/h * 0.9063
West component ≈ 171.98 km/h (rounded to two decimal places)
Now let's consider the wind's velocity, which is given as 45 km/h at 15 degrees north of east. We can again find the North and West components using trigonometry.
The North component of the wind's velocity is given by:
North component = Velocity * sin(Angle)
North component = 45 km/h * sin(15°)
Using a calculator, we find:
North component = 45 km/h * 0.2588
North component ≈ 11.65 km/h (rounded to two decimal places)
The East component of the wind's velocity is given by:
East component = Velocity * cos(Angle)
East component = 45 km/h * cos(15°)
Using a calculator, we find:
East component = 45 km/h * 0.9659
East component ≈ 43.47 km/h (rounded to two decimal places)
Now let's add up the North and West components of both the plane's velocity and the wind's velocity separately.
North component = 80.39 km/h + (-11.65 km/h)
North component ≈ 68.74 km/h (rounded to two decimal places)
West component = 171.98 km/h + (-43.47 km/h)
West component ≈ 128.51 km/h (rounded to two decimal places)
Finally, the plane's new velocity with respect to the ground can be found using Pythagoras' theorem:
New velocity = √(North component^2 + West component^2)
New velocity = √(68.74 km/h^2 + 128.51 km/h^2)
Using a calculator, we find:
New velocity ≈ √(4715.4276 km^2/h^2 + 16522.7601 km^2/h^2)
New velocity ≈ √21238.1877 km^2/h^2
New velocity ≈ 145.73 km/h (rounded to two decimal places)
Therefore, the plane's new velocity with respect to the ground in standard position is approximately 145.73 km/h.