I Keep Getting 2 Different Answers.

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Homework Help Overview

The discussion revolves around determining the resulting velocity of a plane given its speed and the influence of wind. The problem involves vector addition and angle calculations in the context of physics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants describe different methods for calculating the resultant velocity, leading to two distinct angle outcomes. Questions arise regarding the accuracy of specific vector components and the interpretation of wind direction.

Discussion Status

There is ongoing exploration of the problem with participants questioning each other's calculations and assumptions. Some guidance is offered regarding vector representation and angle determination, but no consensus has been reached on the correct values.

Contextual Notes

Participants express confusion over the vector components used in calculations and the implications of wind direction on the resultant velocity. There are references to drawing diagrams and using trigonometric relationships, but no definitive resolution has been provided.

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I Keep Getting 2 Different Answers. Please Help!

1. Plane has a speed of 200 km/h and heads due north; though, 100 km/h northeast wind blows. what is the resulting velocity of plane w/ respect to ground?



2. I keep getting two different answers (I've approached them 2 different ways). I get 147.36 km/h both times but the angle is different! I either get 118.07 degrees and 75.361 degrees. Can anyone confirm one of them?



3. TO get 118.67, I drew the vectors of (-70.71 and 129.29 km/h ---> also tan(x) = 129.29/70.71 ...For the other angle of 75.361, i had tan(x) = 270.711/70.711... BASICALLY I KEEP DRAWING IT DIFFERENTLY :/ Can anyone tell me which ones right?
 
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your this thing "129.29 km/h" seems wrong to me.

I get 270.711

anywyas, how you got 129?
 
thank you! can anyone else confirm?
 
I don't understand where your value of -70.71 comes from? Draw your diagram so you have the 200 km/h in a north direction, and the 100km/h in a southwest direction at the end of the north arrow (a northeasterly wind means a wind blowing from the northeast). Then, the other line in this triangle is the resultant. You can find the angle using the cosine rule for triangles. (NB: the actual angle you want will be the angle of the vector from due north measured in a clockwise way)
 
rootX said:
your this thing "129.29 km/h" seems wrong to me.

I get 270.711

anywyas, how you got 129?

Exactly, i am getting 279.8008km/hr. Take one vector as the plane's velocity and the other vector as the wind's velocity. Now the resultant of this two vectors would be the velocity of the plane w.r.t the ground. And now you can apply the formula as tanx=bsin(theta)/a+bcos(theta).
Here theta is the angle between the two vectors which is 45 and x is the angle which the resultant vector makes with the vector a. Here vector b is the velocity of the plane and vector a is that of wind. Evaluating we get tanx=0.5857. So x is 30.36 this is the angle between the resultant and a. Now a makes an angle of 45 with the positive x axis.So the angle made by the resulatant with the positive x-axis is 75.36
I think that this should be the final answer
 

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