The air speed indicator of a plane that took off from Detroit reads 349 km/h and the compass indicates that it is heading due east to Boston. A steady wind is blowing due north at 46.5 km/h.

a) Calculate the speed of the plane with reference to the ground.

b) If the pilot wishes to fly directly to Boston (due east) what must the compass read? (0 = North, 90 = East, ...)

a) Well, it seems like that plane is facing quite the windy situation! To calculate the speed of the plane with reference to the ground, we need to take into account the wind speed. Since the wind is blowing due north at 46.5 km/h and the plane is heading due east, we can use some good ol' Pythagorean theorem magic.

Let's call the speed of the plane with reference to the ground "x." We can consider the northward component of the wind as 46.5 km/h and the eastward component (the speed of the plane) as x km/h. Therefore, we have a right triangle situation going on.

Using the theorem, we have x² + 46.5² = 349². Time for some math! x² = 349² - 46.5². After crunching the numbers, we get x ≈ 345.5 km/h (rounded to one decimal place).

So, the speed of the plane with reference to the ground is approximately 345.5 km/h. It's braving those winds like a champ!

b) Now, if our brave pilot wants to head directly to Boston (due east), they must consider the wind pushing against them from the north. Since the plane is moving east and the wind is coming from the north, it will create a bit of a crosswind situation.

To fly directly east (0 degrees on the compass), the pilot will need to compensate for the crosswind. So, the compass should read an angle that corresponds to the direction that balances the wind's northward push. In this case, we're looking for the angle that's the arctangent of the wind speed (46.5 km/h) divided by the speed of the plane (345.5 km/h).

But let's be honest, pilots usually have a lot more sophisticated tools to deal with this! So they'll calculate this on their fancy flight computers to make sure they're on the right track.

I hope this helps you navigate the skies with a smile! Safe travels, my friend!

a) To calculate the speed of the plane with reference to the ground, we can use vector addition.

Let's denote the speed of the plane with respect to the ground as Vg and the speed of the wind as Vw.

Since the plane is moving due east and the wind is blowing due north, these two vectors are perpendicular to each other. We can use the Pythagorean theorem to find the resultant velocity:

Vg^2 = Vp^2 + Vw^2

where Vp is the speed of the plane with reference to the air.

Given that Vp = 349 km/h and Vw = 46.5 km/h, we can plug in these values:

Vg^2 = (349 km/h)^2 + (46.5 km/h)^2
Vg^2 = 121,801 km^2/h^2 + 2,152.25 km^2/h^2
Vg^2 = 123,953.25 km^2/h^2

Taking the square root of both sides, we get:

Vg = √(123,953.25 km^2/h^2)
Vg ≈ 352.36 km/h

Therefore, the speed of the plane with reference to the ground is approximately 352.36 km/h.

b) To determine what the compass should read for the pilot to fly directly to Boston (due east), we need to consider the effect of the wind on the plane's heading.

The wind is blowing due north at 46.5 km/h, and the plane is moving with a velocity of 349 km/h due east. If the compass reads θ, then the resultant velocity vector will be:

Vr = Vp + Vw = 349 km/h east + 46.5 km/h north

The angle θ between the resultant velocity vector and the east direction will be equal to the angle between the plane's velocity vector and the east direction minus 90 degrees (since the wind is blowing north).

Let's find the angle between the plane's velocity vector and the east direction:

tan(θ) = Vw / Vp
tan(θ) = 46.5 km/h / 349 km/h
θ = arctan(46.5 km/h / 349 km/h)
θ ≈ 7.51 degrees

Since we want the compass to indicate the actual direction of east, we need to adjust for the wind. Therefore, the compass should read:

θ + 90 degrees = 7.51 degrees + 90 degrees
θ ≈ 97.51 degrees

Therefore, the compass should read approximately 97.51 degrees for the pilot to fly directly to Boston (due east).

To solve this problem, we need to use vector addition to find the resultant velocity of the plane with respect to the ground.

Let's start by visualizing the problem. The plane is moving due east, while the wind is blowing due north. This means that the velocity of the plane with respect to the ground is the vector sum of the airspeed and the wind velocity.

a) To calculate the speed of the plane with reference to the ground, we need to find the magnitude of the resultant vector.

Given:
Air speed of the plane (Vair) = 349 km/h
Wind speed (Vwind) = 46.5 km/h

We can represent the airspeed vector as Vair = 349 km/h due east (along the positive x-axis) and the wind vector as Vwind = 46.5 km/h due north (along the positive y-axis).

Using vector addition, we can find the resultant vector (Vres) by adding these two vectors together. The formula for vector addition is:

Vres = Vair + Vwind

In this case, since the vectors are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant vector:

|Vres| = √(Vair^2 + Vwind^2)

Plugging in the given values, we have:

|Vres| = √(349^2 + 46.5^2) km/h
= √(121801 + 2162.25) km/h
= √124963.25 km/h
≈ 353.39 km/h

Therefore, the speed of the plane with reference to the ground is approximately 353.39 km/h.

b) If the pilot wishes to fly directly to Boston (due east), the compass must read the same angle as the direction of the resultant vector (Vres). In this case, the angle is the arctan(Vwind/Vair), which is the angle between the wind and airspeed vectors.

Using the given values, we can calculate the angle:

θ = arctan(Vwind/Vair)
= arctan(46.5/349)
≈ 7.55 degrees

Therefore, the compass should read approximately 7.55 degrees east of north for the plane to fly directly to Boston.