Calculating Wind Speed for Toronto to Montreal Flight

[chengbin]

Homework Statement

A plane flies 500km from Toronto to Montreal in 2.5h. The return trip takes 3.0h. Assuming the plane flew under constant power, determine:

Homework Equations

a) The speed of the plane in still air

Well that's easy.

b) The speed of the wind. Express your answer rounded to the nearest km/h

The Attempt at a Solution

I don't know what to do. Please help.

 

[HallsofIvy]

If it was easy to get the speed of the plane in still air, its easy to then get the air speed. Since it flew from 500 mi from Toronto to Montreal in 2.5 hours, a "ground" speed of 500/2.5= 200 mph, with the wind assisting, the wind speed is 100 mph minus the speed of the plane in still air. What exactly did you get as the speed of the plane in still air?

 

[tiny-tim]

Welcome to PF!

A plane flies 500km from Toronto to Montreal in 2.5h. The return trip takes 3.0h. Assuming the plane flew under constant power …

Hi chengbin! Welcome to PF!

Hint: Call the velocity of the Plane relative to the Air V_PA

the velocity of the Plane relative to the Ground V_PG

the velocity of the Air relative to the Ground V_AG

and then draw a vector triangle PAG

 

[chengbin]

If it was easy to get the speed of the plane in still air, its easy to then get the air speed. Since it flew from 500 mi from Toronto to Montreal in 2.5 hours, a "ground" speed of 500/2.5= 100 mph, with the wind assisting, the wind speed is 100 mph minus the speed of the plane in still air. What exactly did you get as the speed of the plane in still air?

I was thinking about the same thing. Since 500km/2.5=200km/h, 500/3=166.67km/h, 200-166.67=33.33km/h.

But that doesn't seem to make sense.

Humans can run at around 25-30km/h, so if I'm running at 30km/h and there is a 30km/h coming at me, I won't move at all?

 

[HallsofIvy]

tiny tim, aren't you assuming that the angle the wind makes with the line of flight is known? And that is not given. I think the best you can do is assume the wind direction is along the line from Toronto to Montreal. Strictly speaking what you can find that way is the component of wind speed in that direction. More you cannot find without knowing the wind direction.

Frankly, I have an uncomfortable feeling that when chengbin said it was easy to find the speed of the plane in still air he was only calculating 500/2.5= 200 mph and 500/3= 166 2/3 mph.

chengbin, letting v be the speed of the airplane in still air and letting w be the wind speed (assuming the wind is along the line from Toronto to Montreal, otherwise w is the component of the wind speed in that direction) then the ground speed from Toronto to Montreal is v+ w= 200 and from Montreal to Toronto is v- w= 166 2/3. I don't see how you can calculate one without the other!

 

[tiny-tim]

tiny tim, aren't you assuming that the angle the wind makes with the line of flight is known? And that is not given. I think the best you can do is assume the wind direction is along the line from Toronto to Montreal. Strictly speaking what you can find that way is the component of wind speed in that direction. More you cannot find without knowing the wind direction.

ooh, you're right …

i'm getting an isoceles triangle with a line representing the wind velocity going from the apex to the bae, and dividing the base 6:5 …

but that gives a different answer for each wind direction!