Galilean transformation problem

[Benzoate]

Homework Statement

Pilots are racing small, relatively high-powered airplanes arounds courses marked by a pylon on the ground at each end of the course. Suppose two such evenly matched racers fly at airspeeds of 130 mi/h. Each flies one complete round tripof 25 miles, but their courses are perpendicular to one another and there is a 20mi/h windblowing steadily parallel to one course. (a). Which pilot wins the race and by how much?

Homework Equations

I think this is a Galilean transform'ation problem.
I label the two planes A and B. for plane A , I'll say its velocity u'(A) = u(A)-v , where v is the velocity of the windblowing at 20mi/h and for plane B (u)'(B)=(u)(B). It doesn't really matter but I'll just say the velocity of the wind is parrallel to plane A , but it doesn't really matter.

The Attempt at a Solution

For plane A, u'(A)=u(A)-v -> u'(A)= 130mi/h -20mi/h = 110mi/hr
For plane B , u'(B)=u(B) = 130 mi/hr

In order to determin which plane wins the race by how much I am going to have to determine how much time it will take the planes to arrive at d=25 miles, where d is the total completion of the course
for time A , t(A) = d/u'(A) = (25mi/110mi/hr)*3600 s = 818.18 seconds
for time B , t(B) = d/u'(B) = (25 mi/130mi/hr) * 3600 s = 692.307 seconds
so plane A should arrive at the finish before plane B and plane A wins by delta(t)=t(A)-t(B)= 818.18 secs-692.307 secs =125.87 seconds

but my answer isn't even remotely close to the answer in my textbook. I redid my calculations over and over again and still arrive to the same time intervals. What am I doing wrong?

 

[Doc Al]

Let's make things concrete. Plane A travels along the east-west path; plane B, the north-south track. Let the wind speed be 20 mi/hr east. They travel in straight-line paths to and from the pylons, which are 25/2 = 12.5 mi away from the origin.

Q1: What's the speed of plane A with respect to the ground while traveling east? While traveling west back to the origin?

Q2: What direction must plane B point so that he travels due north? Due south? What is his speed with respect to the ground during each leg of his flight?

 

[Benzoate]

Let's make things concrete. Plane A travels along the east-west path; plane B, the north-south track. Let the wind speed be 20 mi/hr east. They travel in straight-line paths to and from the pylons, which are 25/2 = 12.5 mi away from the origin.

Q1: What's the speed of plane A with respect to the ground while traveling east? While traveling west back to the origin?

Q2: What direction must plane B point so that he travels due north? Due south? What is his speed with respect to the ground during each leg of his flight?

Are my equations correct? I just have the wrong distance
wouldn't the speed of plane A then be 150 mi/h now from plane B perspective and and Plane B velocity will be 110mi/h.

 

[Doc Al]

The speed of plane A is different in each direction; The speed of plane B is not 110 mi/h.

 

[Benzoate]

The speed of plane A is different in each direction; The speed of plane B is not 110 mi/h.

I don't understand. Plane A is only headed in one direction and that's the direction perpendicular to the direction of plane B. Plane A is only headed east-west axis and plane B is headed on the North-south direction. So How can both planes go in many directions? If both planes initially start off at 130mi/hr and I say the wind that's moving at 20mi/hr is headed in the same direction as plane A , then from plane B's perspective the 20 mile/hr wind is speeding up plane A to a velocity of 150 mi/hr. and from plane A's perspective, then plane B's velocity should remain constant.

If all is correct. then , delta(t)=t(B)-t(A) = 12.5mi/130mi/hr - 12.5mi/150mi/hr= 46.153 seconds, meaning plane A should arrive their before plane B by a time difference of 46.153 seconds. Is that calculation correct?

 

[Doc Al]

I don't understand. Plane A is only headed in one direction and that's the direction perpendicular to the direction of plane B. Plane A is only headed east-west axis and plane B is headed on the North-south direction. So How can both planes go in many directions?

The problem says that the planes each travel a roundtrip path of 25 mi. Plane A first travels east for 12.5 mi, then heads back west to return to the starting point. Plane B goes north, then south.

 

[Benzoate]

The problem says that the planes each travel a roundtrip path of 25 mi. Plane A first travels east for 12.5 mi, then heads back west to return to the starting point. Plane B goes north, then south.

I see what you mean. Both planes A and B make a narrow ellispe who's totally perimeter is 25 miles . Okay so , my equations were correct. Whats the point of dividing the diameter if both planes are eventually going to cross the diameter? You still didn't answer my first question. Am I using the right equations or atleast headed in the right direction for this problem?

 

[learningphysics]

I see what you mean. Both planes A and B make a narrow ellispe who's totally perimeter is 25 miles . Okay so , my equations were correct. Whats the point of dividing the diameter if both planes are eventually going to cross the diameter? You still didn't answer my first question. Am I using the right equations or atleast headed in the right direction for this problem?

No, the planes go in straight lines.

Let's look at plane A first (going east/west)... so first it goes eastbound 12.5 miles then it comes back 12.5 miles westbound... What is plane A's speed when it is going east? What is its speed when it is coming back west? Take the wind speed to be east as doc al wrote.

The idea is to add the velocity of the plane relative to the air to the velocity of the wind, to get the total velocity of the plane... direction matters... you're adding vectors. This is especially important for plane B since the wind velocity will not be parallel to the plane velocity... So you add the two vectors to make the plane go in a straight line in the direction you want...

 

[Benzoate]

No, the planes go in straight lines.

Let's look at plane A first (going east/west)... so first it goes eastbound 12.5 miles then it comes back 12.5 miles westbound... What is plane A's speed when it is going east? What is its speed when it is coming back west? Take the wind speed to be east as doc al wrote.

The idea is to add the velocity of the plane relative to the air to the velocity of the wind, to get the total velocity of the plane... direction matters... you're adding vectors. This is especially important for plane B since the wind velocity will not be parallel to the plane velocity... So you add the two vectors to make the plane go in a straight line in the direction you want...

So when plane A is traveling at a velocity of 130mi/hr, there is also the velocity of the wind that's parallel to Plane A who's direction is not going to change. Plane B direction is therefore perpenducular to the direction of the velocity of the wind. If Plane A is headed in the opposite direction compared to the direction of the wind, the from Plane B's position , the velocity of Plane A is will be 110 mi/hr. When plane A changes its course and heads in the same direction as the direction of the wind's velocity, then the velocity of Plane A will be 150mi/hr from Plane B 's vantage point. so would the velocity difference for plane A be : deltaV(A)=V(A)(final) -V(A)(initial)= 150mi/hr-110mi/hr=40mi/hr

Now from plane A's vantage point. the velocity of plane B will initially take off from a south point to a north vantage point with a wind velocity that will tend to try to blow off plane B off course.. the velocity of the wind will always try blow plane B off its course whether plane B is heading from north to south or vice versa. So for both directions of the north bound course, will the velocity of plane B 110 mi/hr?, summing its total velocity to 220 mi/hr?

 

[Doc Al]

I don't understand why you refer to "Plane A's vantage point" versus "Plane B's vantage point". No need to find the relative velocity of the planes, just find their velocities with respect to the ground.

So when plane A is traveling at a velocity of 130mi/hr, there is also the velocity of the wind that's parallel to Plane A who's direction is not going to change.

Plane A's airspeed is 130 mi/hr. The velocity of the air is fixed at 20 mi/hr east with respect to the ground.

Plane B direction is therefore perpenducular to the direction of the velocity of the wind.

Sure. Plane B goes north/south while the wind goes east.

If Plane A is headed in the opposite direction compared to the direction of the wind, the from Plane B's position , the velocity of Plane A is will be 110 mi/hr.

The speed of Plane A will be 110 mi/hr with respect to the ground, not "from Plane B's position".

When plane A changes its course and heads in the same direction as the direction of the wind's velocity, then the velocity of Plane A will be 150mi/hr from Plane B 's vantage point.

Plane A's speed will be 150 mi/hr with respect to the ground.

so would the velocity difference for plane A be : deltaV(A)=V(A)(final) -V(A)(initial)= 150mi/hr-110mi/hr=40mi/hr

Why are you finding the difference in speeds? Find the time it takes for each leg of its journey.

Now from plane A's vantage point. the velocity of plane B will initially take off from a south point to a north vantage point with a wind velocity that will tend to try to blow off plane B off course.. the velocity of the wind will always try blow plane B off its course whether plane B is heading from north to south or vice versa. So for both directions of the north bound course, will the velocity of plane B 110 mi/hr?,

To find plane B's velocity with respect to the ground, you must add his velocity with respect to the air to the velocity of the air. Note that these velocities are vectors and must be added as such. Also: The resultant velocity with respect to the ground must be north/south. (Hint: Draw a right triangle whose hypotenuse is 130.)

summing its total velocity to 220 mi/hr?

Why are you adding velocities for separate legs of the journey? Instead, find the speed and calculate the time of flight. Then you can compare with the time of flight for plane A.

 

[Benzoate]

I don't understand why you refer to "Plane A's vantage point" versus "Plane B's vantage point". No need to find the relative velocity of the planes, just find their velocities with respect to the ground.


Plane A's airspeed is 130 mi/hr. The velocity of the air is fixed at 20 mi/hr east with respect to the ground.

Sure. Plane B goes north/south while the wind goes east.

The speed of Plane A will be 110 mi/hr with respect to the ground, not "from Plane B's position".

Plane A's speed will be 150 mi/hr with respect to the ground.

Why are you finding the difference in speeds? Find the time it takes for each leg of its journey.


To find plane B's velocity with respect to the ground, you must add his velocity with respect to the air to the velocity of the air. Note that these velocities are vectors and must be added as such. Also: The resultant velocity with respect to the ground must be north/south. (Hint: Draw a right triangle whose hypotenuse is 130.)

Why are you adding velocities for separate legs of the journey? Instead, find the speed and calculate the time of flight. Then you can compare with the time of flight for plane A.

I think I finally understand. Here are my equations for the times of A and B

t(A)=L/(130mi/hr +v(river)) +L/((130mi/hr -v(river)) = 12.5mi/(130mi/hr+20mi/hr) + 12.5mi/(130mi/hr -20mi/hr) = .1966 hr = 709.1 seconds

t(B) = L/sqrt((130)^2-(v(river))^2) + L/sqrt((130)^2 -v(river))^2)=2L/sqrt((130)^2-v^2)=25mi/sqrt((130)^2-(20)^2)=.1946 hr=700.649 seconds

making the total time difference to result to this amount: delta(t)=709.1 sec-700.6 sec=8.5 seconds

My numbers may not be exactly correct but is this the general equation I should apply in order to calculate the time values of both planes? I think it is.

 

[Doc Al]

I didn't check your arithmetic, but your equations look good this time. (Assuming by "river" you mean "air". )

 

[Benzoate]

thank you very much for helping me with this problem