Vectors in Physics Question

[collegelife101]

Can someone please help me answer this question?

If an airplane has airspeed of 100 km/hr southwest but is traveling 140 km/hr south relative to the ground, what is the wind velocity?

I guess I'm just not sure how to draw out the vectors. I also didn't quite understand the solution: you must subtract the airspeed (or add its negative) to the resulting ground speed. This makes for a 1-1-square root of 2 triangle, so the wind must also be 100 km/hr.

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[NextStepTutor_1]

Hi,

In this case, the problem is stating that there is a plane with a velocity of 100 km/hr southwest, however, the resultant vector of its motion (the velocity relative to the ground) is 140 km/hr south. Thus, we are charged with finding the second velocity vector (wind velocity in this case) which when added to the original speed of the plane gives the total velocity.

Now, the plane has a velocity vector pointing southwest, and the resultant vector points south, so the difference in angle of 45 degrees. This is what makes the triangle a 45-45-90 triangle. To get the wind vector we use Pythagoras theorem to find the missing value of wind = 100 km/hr.

 

[collegelife101]

Thank you for your response, but how did you know that the difference in angle is 45 degrees?

Hi,

In this case, the problem is stating that there is a plane with a velocity of 100 km/hr southwest, however, the resultant vector of its motion (the velocity relative to the ground) is 140 km/hr south. Thus, we are charged with finding the second velocity vector (wind velocity in this case) which when added to the original speed of the plane gives the total velocity.

Now, the plane has a velocity vector pointing southwest, and the resultant vector points south, so the difference in angle of 45 degrees. This is what makes the triangle a 45-45-90 triangle. To get the wind vector we use Pythagoras theorem to find the missing value of wind = 100 km/hr.

 

[syoung]

Draw a right-angled triangle. Label the hypotenuse 140 and one leg 100. Use trig to find the angle. Should be close to 45.

 

[NextStepTutor_1]

You could find the angles for the other sides by using law of sines/cosines, however, that isn't what the MCAT covers in terms of math. So, I'm wondering if this was an MCAT question with answer choices--if so, they might have expected you to plug in answer choices and use process of elimination to see which answers are wrong and then choose the correct choice.