# Physics plane speed problem

I want to know if my steps taken are correct...
Thanks...

A plane has an airspeed of 235km/h. It takes off and wants to fly to a city that is directly north of it at a distance of 455km. The wind is blowing to the east east at 42km/h. Find a) The course that the pilot must steer in order to fly directly north to the city and b) the time it will take to arrive at the city.
For a, I broke it up into the vector components; 42i + 235j and took the absolute value of it to get 238.72km/h.... and stated that the plane must go at this speed to the northeast.
For b) I used the formula s = ut and did 455 = 238.72*t and solved for t.

Are my methods correct or did I miss something? Thanks.

LeonhardEuler
Gold Member
Pseudo Statistic said:
I want to know if my steps taken are correct...
Thanks...

A plane has an airspeed of 235km/h. It takes off and wants to fly to a city that is directly north of it at a distance of 455km. The wind is blowing to the east east at 42km/h. Find a) The course that the pilot must steer in order to fly directly north to the city and b) the time it will take to arrive at the city.
For a, I broke it up into the vector components; 42i + 235j and took the absolute value of it to get 238.72km/h.... and stated that the plane must go at this speed to the northeast.
For b) I used the formula s = ut and did 455 = 238.72*t and solved for t.

Are my methods correct or did I miss something? Thanks.
If the wind is blowing to the east and the pilot heads northeast, then the plane will end up going even farther east than if it just headed north. Also, "northeast" probably is not specific enough. The way to solve the problem is to say, ok the vector j points north and the vector i points east. Now if I want the plane to head north, then the component of the plane's speed to the east is zero. So the plane should head a little west so that its western component exactly cancels the eastern component. You know that the magnitude of the plane's velocity is 235km/h. Split it into components and set the western component of the planes speed equal to the eastern component due to the wind so that they cancel.

:|...
I'm lost when you say split it up into components because I wasn't given any angles or anything...
I made a mistake, I meant it would have to go 238.72km/h to the north WEST in order to overcome this interference; i.e. it has to travel west at 42km/h....
I'm confused about what you mean.

Doc Al
Mentor
Consider this vector equation:
$\vec{V}_{p/g} = \vec{V}_{p/a} + \vec{V}_{a/g}$

The east-west component of the plane's air speed (plane with respect to air; p/a) must exactly cancel the air's eastward speed. Use that fact to find the angle that the plane must fly with respect to the air.

Alright, thanks.
From what you're saying, the angle = arctan (235/-42)? (Sorry for my ignorance)
And my answer for part B... is it correct?

Doc Al
Mentor
Pseudo Statistic said:
From what you're saying, the angle = arctan (235/-42)?
No. Think this way: The plane's airspeed is fixed at 235 km/hour. But it's east-west component (its $\hat{i}$ component) must equal 42 km/hour west. What angle must the plane's velocity (with respect the air) make to have an x-component of - 42 km/hour?

And my answer for part B... is it correct?
You can't do part B until you find the plane's actual speed with respect to the ground. (That's the north-south component of the airspeed.)