- #1
fara0815
- 45
- 0
Hello there!
First, I really want to thank you guys for helping me out with my physics problems. I would not have been able to make such a good progress I have done in the last couples of days!
I started a new chapter, it is about circula motion and I am having a hard time with this one:
A drunken student fell as sleep on a bench. When he is finally waking up, he can see a street light which is 50 m away from him. He is getting up and is orbiting the street light with a constant angular speed of [tex] 0,03 s^{-1} [/tex] on a helix curve which is getting tighter and tighter. The helix curve has the form of
[tex] s_\varphi=(R-k\varphi)(\cos\varphi \\ \sin\varphi) [/tex] (the cos and sin should be vektor components for x and y but I do not know how to do that with latex)
Where R is the distance between the bench and the street light, [tex] \varphi [/tex] is the angle between the direct distance of the student and the street light and k is a constant of 2m.
How great is the distance between the student and the bench after 6 min he stood up ?Here are my thoughts:
To get the angle I multiply the angular speed with the time:
[tex] \varphi= \omegat = 0,03(6*60) = 10.8 [/tex]
I thought using that angle to get the x and y components of the student's position (zero of the coordinate system is in the center of the street light) with which I can get the modulus would be the solution but I never got the 62,116 m.
I am aware that I have to consider the distance of the bench to the street light as well to get the over all distance between the bench and the student.
Does anyone out there have a hint for me?
Many thanks!
First, I really want to thank you guys for helping me out with my physics problems. I would not have been able to make such a good progress I have done in the last couples of days!
I started a new chapter, it is about circula motion and I am having a hard time with this one:
A drunken student fell as sleep on a bench. When he is finally waking up, he can see a street light which is 50 m away from him. He is getting up and is orbiting the street light with a constant angular speed of [tex] 0,03 s^{-1} [/tex] on a helix curve which is getting tighter and tighter. The helix curve has the form of
[tex] s_\varphi=(R-k\varphi)(\cos\varphi \\ \sin\varphi) [/tex] (the cos and sin should be vektor components for x and y but I do not know how to do that with latex)
Where R is the distance between the bench and the street light, [tex] \varphi [/tex] is the angle between the direct distance of the student and the street light and k is a constant of 2m.
How great is the distance between the student and the bench after 6 min he stood up ?Here are my thoughts:
To get the angle I multiply the angular speed with the time:
[tex] \varphi= \omegat = 0,03(6*60) = 10.8 [/tex]
I thought using that angle to get the x and y components of the student's position (zero of the coordinate system is in the center of the street light) with which I can get the modulus would be the solution but I never got the 62,116 m.
I am aware that I have to consider the distance of the bench to the street light as well to get the over all distance between the bench and the student.
Does anyone out there have a hint for me?
Many thanks!
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