How Does Wind Direction Affect Perception of Speed in Moving Vehicles?

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Wind direction perception in moving vehicles can be counterintuitive; as a motorcar increases speed from 4m/s to 6m/s while heading west, the driver feels the wind from the southeast instead of the southwest. This discrepancy raises questions about the relationship between vehicle speed and wind perception. In a separate scenario involving a hunter chasing a dog, the discussion highlights that the acceleration of the hunter is not purely centripetal, as it involves changes in direction rather than circular motion. The participants suggest analyzing the problem with small time intervals to understand the dynamics better. Overall, the conversation emphasizes the complexities of motion and perception in physics.
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Q1:
There is a man riding a motorcar, and it move toward west with 4m/s, and the man feel the wind blow from the south. If the motorcar increase the speed to 6m/s, he feel the wind is blow from SE(or135), so what is the wind speed?

When I do this question, I don't know why the man can feel the wind blow from SE when he increase his speed, isn't suppose he feels the wind blow from SW??

Q2:
[URL=http://www.dumpt.com/img/viewer.php?file=9e51a6ndf3axbrdjvgp2.png][PLAIN]http://www.dumpt.com/img/files/9e51a6ndf3axbrdjvgp2_thumb.png[/URL][/PLAIN]

There is a dog, move on the upper line with a constant speed V1. and a hunter chase the dog with a speed v2, and it's motion is always toward the dog(point toward). At a pt. D, and F, where FD is perpendicular to the upper line. Find the acc. of the hunter. (where FD = L)

When I am doing this question, I only know it is related to circular motion, but when I saw the answer, I had no idea what is going on. the answer is v1xv2/L. Could anyone tell me is that there are only a Centripetal acc?





(I am come from hong kong, forgive my poor english)
 
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henry407 said:
Q1:
There is a man riding a motorcar, and it move toward west with 4m/s, and the man feel the wind blow from the south. If the motorcar increase the speed to 6m/s, he feel the wind is blow from SE(or135), so what is the wind speed?

When I do this question, I don't know why the man can feel the wind blow from SE when he increase his speed, isn't suppose he feels the wind blow from SW??

Q2:
[URL=http://www.dumpt.com/img/viewer.php?file=9e51a6ndf3axbrdjvgp2.png][PLAIN]http://www.dumpt.com/img/files/9e51a6ndf3axbrdjvgp2_thumb.png[/URL][/PLAIN]

There is a dog, move on the upper line with a constant speed V1. and a hunter chase the dog with a speed v2, and it's motion is always toward the dog(point toward). At a pt. D, and F, where FD is perpendicular to the upper line. Find the acc. of the hunter. (where FD = L)

When I am doing this question, I only know it is related to circular motion, but when I saw the answer, I had no idea what is going on. the answer is v1xv2/L. Could anyone tell me is that there are only a Centripetal acc?


(I am come from hong kong, forgive my poor english)

Q1. I agree with your concern - either the new wind is SW or the car motion is East.

Q2. I don't think this is circular motion. The acceleration will be calculated from limits of changes in velocity of the hunter - due to changes of course only in direction.
To get a feel for this, assume some values, and see what is happening. Consider the situation given, then what we would have a small time interval before and after.
perhaps make that time interval 0.1 second at first, then 0.01 seconds and see what is happening.
Also, presumable v2 is greater than v1, or the hunter is never going to catch the dog.
Don't lose sight of the idea that for very small angles, TanA, A and SinA are all approximately equal - A measured in radians [not degrees]
 
PeterO said:
Q1. I agree with your concern - either the new wind is SW or the car motion is East.

Q2. I don't think this is circular motion. The acceleration will be calculated from limits of changes in velocity of the hunter - due to changes of course only in direction.
To get a feel for this, assume some values, and see what is happening. Consider the situation given, then what we would have a small time interval before and after.
perhaps make that time interval 0.1 second at first, then 0.01 seconds and see what is happening.
Also, presumable v2 is greater than v1, or the hunter is never going to catch the dog.
Don't lose sight of the idea that for very small angles, TanA, A and SinA are all approximately equal - A measured in radians [not degrees]

Thinks for ur opinion first. I still think it is a circular motion, isn't it (the only thing is the radiu is keep changing)., in the question, we can observe that the angle made by the hunter and the dog is 90 degree(I don't like radian, althought all my physics teacher tell me to use radian).
o, after I read the question, I had a thought of using a= v^(2)/R. But it only can calculate the centirpal acc., but there are also tangent acc.(the name maybe wrong).
 
henry407 said:
Thinks for ur opinion first. I still think it is a circular motion, isn't it (the only thing is the radiu is keep changing)., in the question, we can observe that the angle made by the hunter and the dog is 90 degree(I don't like radian, althought all my physics teacher tell me to use radian).
o, after I read the question, I had a thought of using a= v^(2)/R. But it only can calculate the centirpal acc., but there are also tangent acc.(the name maybe wrong).

Firstly, every curve can be approximated to a circle, if you look at a small enough section of the curve.
That is born out by the fact that a small spherical mirror will focus light almost as well as a parabolic mirror.

If the hunter was traveling in a circle, the centre of that circle would be somewhere to the right of the point where he currently is, and why you would choose the radius of that circle to be the same as the distance to the dog I am not sure ?

There is no tangential acceleration, so v^2 / R might be useful, but you do not know the value of R anyway?
Remember, R is the distance to the centre of the circle - and you only know that the centre of the circle is to the right, not how far away it is.
 
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