Nancy is a pilot in Canada¡¯s North. She needs to deliver emergency supplies to a location that is 500 km away.
Nancy has set the aircraft¡¯s velocity to 270 km/h on a northbound heading. The wind velocity is 45 km/h from the
east. Determine the resultant ground velocity of the aircraft.
To determine the resultant ground velocity of the aircraft, we need to take into account the velocity of the aircraft and the velocity of the wind.
Since the aircraft is heading in a northbound direction, we can break down its velocity into two components: one in the north direction and one in the east direction.
The north component of the aircraft's velocity is 270 km/h, and the east component of the wind velocity is 45 km/h.
To calculate the resulting north component of the aircraft's velocity, we subtract the east component of the wind velocity from the north component of the aircraft's velocity:
North component of resultant velocity = 270 km/h - 45 km/h = 225 km/h
The east component of the aircraft's velocity is equal to the east component of the wind velocity:
East component of resultant velocity = 45 km/h
So, the resultant ground velocity of the aircraft is a velocity of 225 km/h in the north direction and 45 km/h in the east direction.
To determine the resultant ground velocity of the aircraft, we need to consider the effect of both the aircraft's velocity and the wind velocity. To do this, we will use vector addition.
Step 1: Convert the velocities into vector form.
The aircraft's velocity is given as 270 km/h on a northbound heading. Since it is heading north, the velocity vector would point directly north (upwards). So, the aircraft's velocity vector would be 270 km/h north (upwards) or 270 km/h in the positive y-direction.
The wind velocity is given as 45 km/h from the east. Since it is coming from the east, the velocity vector would point directly west (left). So, the wind velocity vector would be 45 km/h west (left) or 45 km/h in the negative x-direction.
Step 2: Add the velocity vectors.
To add the velocity vectors, we can simply add their corresponding components. In this case, we add the x-components and the y-components separately.
For the x-components:
The aircraft's x-component is 0 km/h since it's not moving east-west.
The wind's x-component is -45 km/h since it's coming from the east (opposite to the positive x-direction).
Adding these two x-components, we get 0 km/h + (-45 km/h) = -45 km/h.
For the y-components:
The aircraft's y-component is 270 km/h since it's moving north (in the positive y-direction).
The wind's y-component is 0 km/h since it's not affecting the north-south movement.
Adding the two y-components, we get 270 km/h + 0 km/h = 270 km/h.
So, the resultant velocity is -45 km/h in the x-direction (west) and 270 km/h in the y-direction (north).
Step 3: Calculate the magnitude and direction of the resultant velocity.
To calculate the magnitude of the resultant velocity, we can use the Pythagorean theorem.
Magnitude = sqrt((-45 km/h)^2 + (270 km/h)^2)
Magnitude = sqrt(2025 km^2/h^2 + 72900 km^2/h^2)
Magnitude = sqrt(74925 km^2/h^2)
Magnitude ≈ 273.67 km/h
To determine the direction of the resultant velocity, we can use trigonometry.
tan(θ) = opposite/adjacent
tan(θ) = 270 km/h / -45 km/h
θ ≈ -81.87 degrees
Since the angle is measured clockwise from the positive x-axis, we can say that the resultant ground velocity is approximately 273.67 km/h at an angle of 81.87 degrees west of north.
same type as your previous question
resultant vector = 270(cos90, sin90) + 45(cos180,sin18)
= ....
then find the magnitude