Find th resultant and speed direction of the airplane relative to the ground....

AN airplane is heading west at velocity 950km/hr and wind is blowing northward at velocity of 55.o km/hr

This only makes sense if the 950 km/h west velocity is the velocity of the plane relative to the air

Add it vectorially to the wind velocity vector, to get the velocity of the plane with respect to the ground.

It will have components of 950 km/h west and 55 km/h north.

The vector sum has magnitude 951.6 km/h. You can get the direction from the ratio of components (arctan 55/950 north of west)

To find the resultant velocity and speed direction of the airplane relative to the ground, we need to use vector addition.

Step 1: Assign coordinate axes to the scenario. Let's assume the positive x-axis is east, the positive y-axis is north, and the origin is the starting point of the airplane.

Step 2: Draw a vector representing the velocity of the airplane. Since the airplane is heading west (opposite to the positive x-axis) with a velocity of 950 km/hr, the velocity vector would point towards the west with a magnitude of 950 km/hr.

Step 3: Draw a vector representing the wind blowing northward. Since the wind is blowing northward (in the positive y-axis direction) with a velocity of 55.0 km/hr, the wind vector would point towards the positive y-axis with a magnitude of 55.0 km/hr.

Step 4: Use vector addition to find the resultant velocity. Add the airplane's velocity vector to the wind vector (since they are acting in different directions).

To do this, we can break down the velocities into their x and y component vectors using trigonometry:

Velocity of the airplane in the x-direction = -950 km/hr (since it's heading west)
Velocity of the airplane in the y-direction = 0 km/hr (no component in the north direction)
Velocity of the wind in the x-direction = 0 km/hr (no component in the east-west direction)
Velocity of the wind in the y-direction = 55.0 km/hr

So, the resultant velocity vector in the x-direction is -950 km/hr and in the y-direction, it is 55.0 km/hr.

Step 5: Find the magnitude and direction of the resultant velocity vector. To find the magnitude (speed), we can use the Pythagorean theorem:

Resultant velocity (speed) = sqrt((-950)^2 + 55.0^2) km/hr = sqrt(902,500 + 3,025) km/hr = sqrt(905,525) km/hr ≈ 951.9 km/hr

To find the direction, we can use the inverse tangent function:

Resultant velocity direction = arctan(55.0 / 950) = 3.18° north of west (rounded to two decimal places)

Therefore, the resultant velocity (speed direction) of the airplane relative to the ground is approximately 951.9 km/hr, directed 3.18° north of west.

To find the resultant velocity and direction of the airplane relative to the ground, we need to use vector addition.

Step 1: Represent the airplane's velocity as a vector. The airplane is heading west, so its velocity vector points in the west direction and has a magnitude of 950 km/hr.

Step 2: Represent the wind's velocity as a vector. The wind is blowing northward, so its velocity vector points in the north direction and has a magnitude of 55.0 km/hr.

Step 3: Add the two velocity vectors together using vector addition. Since the airplane is heading west and the wind is blowing northward, we can add the vectors by placing them head-to-tail. The resultant vector represents the airplane's velocity relative to the ground.

Step 4: To add the vectors, draw the velocity vector of the airplane (950 km/hr) pointing west. From the tip of this vector, draw the velocity vector of the wind (55.0 km/hr) pointing north.

Step 5: Complete the parallelogram by drawing a line from the tail of the airplane's velocity vector to the tip of the wind's velocity vector.

Step 6: Measure the magnitude and direction of the resultant vector. The magnitude can be measured using a ruler, and the direction can be determined by measuring the angle it makes with the west direction (or by using trigonometry).

Once you have determined the magnitude and direction of the resultant vector, you will have the speed and direction of the airplane relative to the ground.