I need help on solving this problem
A plane with an air speed of 400km/hr wants to go north but a wind of 70 km/hr is blowing west. What must be the planes heading (to go north)? What will be its resulting ground speed?
To solve this problem, we can use vector addition.
Step 1: Draw a diagram representing the problem. Draw a line to represent the plane's heading to the north, and another line to represent the wind blowing to the west.
Step 2: Define the vectors:
- Let's call the velocity of the plane "Vp" and the velocity of the wind "Vw".
- The magnitude of Vp (air speed) is 400 km/hr, and its direction is the desired heading to the north.
- The magnitude of Vw (wind speed) is 70 km/hr, and its direction is west.
Step 3: Determine the resulting velocity of the plane. This can be found by adding the vectors Vp and Vw.
Step 4: Use the Pythagorean theorem to find the magnitude of the resulting velocity (ground speed). The Pythagorean theorem states: c^2 = a^2 + b^2, where c is the hypotenuse of a right triangle, and a and b are the other two sides. In this case, a is the magnitude of Vp, and b is the magnitude of Vw.
Step 5: Use the inverse tangent function to find the angle between the resulting velocity and the positive x-axis. This angle will give the plane's heading.
Let's calculate the plane's heading and its resulting ground speed:
Step 1: Draw a diagram representing the problem.
–––> Vp (400 km/hr)
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|
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V
Vw (70 km/hr)
Step 2: Define the vectors:
- Vp = 400 km/hr to the north
- Vw = 70 km/hr to the west
Step 3: Determine the resulting velocity of the plane.
- Vp + Vw = sqrt((400^2) + (70^2)) km/hr ≈ 408.25 km/hr
- The resulting velocity is directed at an angle with respect to the positive x-axis.
Step 4: Use the Pythagorean theorem to find the magnitude of the resulting velocity (ground speed).
- Ground Speed ≈ 408.25 km/hr
Step 5: Use the inverse tangent function to find the angle between the resulting velocity and the positive x-axis. This angle will give the plane's heading.
- tan^(-1)(70 / 400) ≈ 9.9 degrees
- The plane's heading is approximately 9.9 degrees north of east.
Therefore, the plane's heading to go north should be approximately 9.9 degrees north of east, and its resulting ground speed will be approximately 408.25 km/hr.
To solve this problem, we can use vector addition. Let's break down the problem step by step:
Step 1: The plane's air speed is given as 400 km/hr. This means that in calm air, the plane can travel 400 km in one hour.
Step 2: The wind is blowing at a speed of 70 km/hr towards the west. Since the plane wants to go north, we can consider the wind as a vector acting in the west direction.
Step 3: To find the plane's heading (to go north), we need to find the resultant vector by adding the vectors of the plane's air speed and the wind. Since the plane is going north and the wind is blowing west, we need to find the vector sum of these two vectors.
Step 4: To find the resultant vector, we can use the Pythagorean theorem. The magnitude of the resultant vector is given by the square root of the sum of the squares of the magnitudes of the individual vectors.
Let's calculate the resultant vector:
Magnitude of the resultant vector = √(400^2 + 70^2) = √(160000 + 4900) = √164900 ≈ 406.12 km/hr
Step 5: The direction of the resultant vector can be found using trigonometry. We can use the inverse tangent (arctan) function to find the angle of the resultant vector in relation to north. The formula is:
Angle = arctan (opposite/adjacent)
In this case, the opposite side is the wind vector (70 km/hr) and the adjacent side is the plane's air speed (400 km/hr). Therefore, the angle is:
Angle = arctan (70/400) ≈ 9.96 degrees
So, the plane's heading to go north is approximately 9.96 degrees west of north.
Step 6: The resulting ground speed is nothing but the magnitude of the resultant vector, which is approximately 406.12 km/hr. This means that the plane will be traveling at a speed of 406.12 km/hr relative to the ground.
To summarize:
- The plane's heading (to go north) is approximately 9.96 degrees west of north.
- The resulting ground speed is approximately 406.12 km/hr.
The plane must have an air speed component of 70 km/hr east to compensate for the 70 km/h wind speed west. That will leave a north component of air speed of sqrt[400^2 - 70^2] = 393.8 km/h
Required heading = arcsin 70/400 = 10.1 degrees (E of N)