A plane leaves Toronto and flies with an airspeed of 2.20 x 10^2 km/h always pointing due east. A wind is blowing from the north at 8.0 x 10^1 km/h.

a) What is the plane's velocity relative to the ground?
b) What is the plane's displacement from Toronto after flying for 2.5h?

a) I did the pythagorean theorem and got 234 km/h. I did Tan O = O/A and got [E20N] for my direction.

b) I got 585 km.

Did I get the right answers? I was a bit confused by part a because I thought it also involved gravity.

For (a), draw and solve a vector diagram.

Air velocity + Airspeed vector = Ground speed vector
For (b) Multiply the Ground speed vector by 2 hours, and the answer will be the displacement vector.
I will be glad to critique your work.

To find the plane's velocity relative to the ground, we need to combine the effect of the plane's airspeed and the wind.

a) To calculate the plane's velocity relative to the ground, we can use vector addition. The airspeed acts eastward, and the wind acts northward. Let's break down the velocities into their horizontal and vertical components.

The plane's airspeed has a magnitude of 2.20 x 10^2 km/h and is directed due east. So, its horizontal component is 2.20 x 10^2 km/h.

The wind has a magnitude of 8.0 x 10^1 km/h and is directed north. So, its vertical component is 8.0 x 10^1 km/h.

To find the plane's velocity relative to the ground, we need to add these components. Since the horizontal and vertical components are perpendicular, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:

Resultant velocity = √(horizontal component)^2 + (vertical component)^2

Resultant velocity = √(2.20 x 10^2 km/h)^2 + (8.0 x 10^1 km/h)^2

Using a calculator, the resultant velocity is approximately 2.34 x 10^2 km/h.

b) To find the plane's displacement after flying for 2.5 hours, we need to multiply its velocity relative to the ground by the time.

Displacement = velocity × time

Displacement = 2.34 x 10^2 km/h × 2.5 h

Using a calculator, the displacement is approximately 5.85 x 10^2 km.

Therefore, after flying for 2.5 hours, the plane's displacement from Toronto is approximately 5.85 x 10^2 km.

To find the plane's velocity relative to the ground, we need to consider the vector addition of the plane's airspeed and the wind velocity.

a) The airspeed of the plane is given as 2.20 x 10^2 km/h due east. Since it is always pointing east, we can represent this velocity as a vector with magnitude 2.20 x 10^2 km/h and an angle of 0 degrees.

The wind velocity is given as 8.0 x 10^1 km/h from the north. Since it is blowing from the north, we can represent this velocity as a vector with magnitude 8.0 x 10^1 km/h and an angle of 90 degrees.

To find the plane's velocity relative to the ground, we can add these vectors using vector addition. In this case, since the vectors are perpendicular, we can use the Pythagorean theorem to find the magnitude of the resulting vector and trigonometry to find its angle.

Magnitude of resultant vector = sqrt((2.20 x 10^2)^2 + (8.0 x 10^1)^2) ≈ 233.17 km/h

Angle of resultant vector = atan((8.0 x 10^1) / (2.20 x 10^2)) ≈ 20.64 degrees (measured counterclockwise from east)

Therefore, the plane's velocity relative to the ground is approximately 233.17 km/h at an angle of approximately 20.64 degrees counterclockwise from due east.

b) To find the plane's displacement from Toronto after flying for 2.5 hours, we need to multiply the plane's velocity relative to the ground by the time.

Displacement = Velocity x Time = 233.17 km/h x 2.5 h

Therefore, the plane's displacement from Toronto after flying for 2.5 hours is approximately 582.93 km in the direction obtained in part (a), which is approximately 20.64 degrees counterclockwise from due east.