Help on Kinematics: Trip Time Calculation w/ 60km/h Wind

[ellusion]

A pilot flies with a constant speed of 120km/h relative to the air, and makes instantaneous turns, when necessary. He follows a perfectly square path on the ground, using north-south and east-west roads as a guide for each of the 60km sides. On a daywhen there is a steady 60km/h wind blwoing diagonally across the square(northeast), how long does the trip take?

I drew a picture and found the distance of the diagonal, from there i am just clueless on how to determine time of the trip. any suggestions?

 

[geoffjb]

A pilot flies with a constant speed of 120km/h relative to the air, and makes instantaneous turns, when necessary. He follows a perfectly square path on the ground, using north-south and east-west roads as a guide for each of the 60km sides. On a daywhen there is a steady 60km/h wind blwoing diagonally across the square(northeast), how long does the trip take?

I drew a picture and found the distance of the diagonal, from there i am just clueless on how to determine time of the trip. any suggestions?

Determine the x- and y-components of the wind, and add them to the velocities of each of the pilot's legs. Determine how long each leg takes, and add them together.

 

[drpizza]

There are a variety of methods to solve this problem... perhaps the simplest to understand is to work on each leg of the trip separately.

Here are a couple things to think about: If I fly due north with a speed of 100mph, relative to the air, and the air is blowing 20mph due north; my net speed, compared to the ground, is 120 mph due north.

Suppose I point my plane due North and have a speed 100mph relative to the air and the air is blowing 20 mph due East. Compared to the ground, I won't actually be moving due North; the wind will cause me to drift Eastward at 20mph. To find my net speed, I'd use the pythagorean theorem.

Now, if I wanted to fly due North, relative to the ground, I'd need a westward component of my velocity to "cancel out" the Eastward drift due to the wind. I'd have to point my plane somewhere west of north. 100mph becomes the hypotenuse, 20 mph west is the horizontal component, and I'd have to find the vertical component. (I can also find the angle to determine which way to actually point my plane using basic right triangle trig.) That vertical component will tell me how fast I end up moving relative to the ground.

Now, what you can do to solve your problem is to break the wind into perpendicular components and combine the two things I did above for each leg of the trip.

Maybe someone else has a better idea on a method which helps you understand the problem, and gets to a solution quicker.

 

[ellusion]

ok i found the x and y components they are both 42.43m/s. what do you mean by leg?

 

[geoffjb]

ok i found the x and y components they are both 42.43m/s. what do you mean by leg?

A "leg" to a pilot is basically each component of the trip. In this case, there are four legs: North, East, South, and West.

 

[ellusion]

So for the north it would be 120km/h + 42.43km/h = 162.43km/h
t = d/t = 60/162.43 = 0.37h?

 

[geoffjb]

So for the north it would be 120km/h + 42.43km/h = 162.43km/h
t = d/t = 60/162.43 = 0.37h?

If he's following roads (the ground), is he pointing his plane along the road?

That is, is his full velocity directed along the road to keep him aligned with it?

 

[ellusion]

when add all the legs together i get 1.48h. the answer is 2.5h
whats wrong here.

 

[drpizza]

Correct me if I'm wrong, but I believe Geoff's solution is mistaken; if you simply add the vectors, you won't be going "due North". I believe the problem has a more direct approach, using law of cosines - draw a vertical line pointing North. From the bottom of this line, draw a vector northeast, 60 km/h long (to scale). From the end of this, draw a vector with a length of 120km/h (to scale) which returns to your due north heading. You have a triangle. The North component is unknown; the bottom angle is 45 degrees, and you have the length of two sides. Apply the law of cosines, and you directly have the magnitude of the North component. Unless you're really familiar with the law of cosines, this probably makes less sense than breaking it into separate vectors.

edit: I see Geoff has now addressed this with a later question about whether all of his velocity is directed north in order to maintain a due north heading.

 

[ellusion]

he has a constant velocity of 120km/h but with wind affecting him he's not aligned?

 

[radou]

\vec{v}_{pilot}=\vec{v}_{air}+\vec{v}_{pilot,air}. Does this require so much philosophy and different approaches?

You know the velocity of the wind (air), as you do know the velocity of the pilot relative to the air.

 

[drpizza]

Hmmm.

\vec{v}_{pilot}=\vec{v}_{air}+\vec{v}_{pilot,air}. Does this require so much philosophy and different approaches?

You know the velocity of the wind (air), as you do know the velocity of the pilot relative to the air.

Yes, you know the velocity of the wind. But you do NOT know the velocity of the pilot relative to the air. You only know the speed of the pilot relative to the air, but not the direction. He is not heading due north (for the first leg) at 120km/hr, relative to the air. If he were, his actual track would not be due north relative to the roads on the ground.

Working on the problem one leg of the journey at a time, I can only think of two methods - breaking everything up into N/S vectors and E/W vectors or using the law of cosines.