Vector expressions - equivalence

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A car traveling at 12 m/s experiences wind at 8 m/s perpendicular to the road, leading to confusion about the wind's velocity relative to the ground. Two vector expressions are proposed: one adds the car's velocity to the wind's velocity relative to the car, while the other subtracts the car's velocity from the wind's velocity relative to the ground. Both expressions yield the same wind magnitude of approximately 14 m/s, but the direction differs due to the sign convention. The discussion highlights the importance of understanding vector directionality and the potential for misinterpretation when visualizing real-world scenarios. Confidence in vector arithmetic is emphasized as a key aspect of resolving such problems.
jemerlia
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Homework Statement


A car is traveling at 12ms^-1. To the passenger in the car the wind appears to be blowing at 8.0ms^-1 at right angles to the road. What is the magnitude and direction of the velocity of the wind with respect to the ground.


Homework Equations


I can think of two possible vector expressions which should be equivalent. I am clearly doing something wrong because they are not.
Using:
vwg = velocity of wind wrt ground
vcg = velocity of car wrt ground
vwc = velocity of wind wrt car
Expression 1
vwg = vwc + vcg

Expression 2 (following the rule of subtracting the observer's movement)
vwc= vwg-vcg

Yes - I'm aware I should add the negative vector and that the negative sign means the reverse direction

The Attempt at a Solution



Expression 1:
The solution vector diagram is (excuse the dots so I don't lose the spaces):

^----------->vcg
|.....^
|vwc ... /
|... / vwg
|.../
|../
|/
It gives a vwg magnitude of c. 14ms^-1 and a direction of tan^-1(8/12) = c. 34 degrees with the wind coming from behind the car

Expression 2:
The solution vector diagram is:

<----------------- -vcg
.^......^
...\......|
...\ ...| vwc
...vwg..\....|
.....\...|
.....\...|
......\..|
.....\|
Of course the magnitude is the same as in the previous example but the direction is
tan^-1 (8/-12) = -34 degrees.

My concern is that the wind appears to be coming from the front!

Gloom - what have I misunderstood?
 
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Your method is correct. The answer is not exactly 14m/s. In both cases it is only the direction you start with that results in the -ive or +ive sign. Why do you think the wind appears to be coming from the front?
 
Thank you for your reassurance. I lacked the confidence to be certain the second vector diagram was the equivalent of the first - especially because vector arithmetic is new to me. I would like to ask this naive question - if a, b and c are vector quantities and they are related:
a = b+c
then is it correct to transpose so that:
b= a-c
?

Why did I think the wind was coming from the front? The honest answer is twofold: first: sometimes I have a pathological inability to imagine the behaviour of the real word and, second I had convinced myself the second expression/vector diagram was telling a different story to the first.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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