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Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

 

 

I need help!!!!!

 Oct 31, 2019
 #1
avatar+236 
+1

There is a reason as to why no one has answered your question, me included. First, you have posted this problem twice. Second, the use of words in your title, HELP PLZ HELP suggest some very urgent help. Why would you need urgent help if you're not cheating. Because of this, most Web2.0calc users will assume that you are cheating. Lastly, you have, frankly, quite a rude message at the bottom of your question, " I need help!!!!! ". Notice the use of exclamation marks. To be honest, there's no reason for us to answer your question.

 Nov 1, 2019
 #2
avatar+8652 
+1

Odell and Kershaw run for 30 minutes on a circular track. Odell(1) runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw(2) runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they meet?

 

Odell und Kershaw rennen 30 Minuten lang auf einer Kreisbahn. Odell(1) fährt mit 250 m / min im Uhrzeigersinn und nutzt die innere Fahrspur mit einem Radius von 50 Metern. Kershaw(2) fährt mit 300 m / min gegen den Uhrzeigersinn und benutzt die äußere Fahrspur mit einem Radius von 60 Metern, beginnend auf der gleichen radialen Linie wie Odell. Wie oft treffen sie sich nach dem Start?

 

Hello Guest!

 

\(\alpha= \omega t=\frac{v}{r}\cdot t\\ \alpha_1= \omega_1 t=\frac{v_1}{r_1}\cdot t\\ \alpha_2=2\pi - \omega_2 t=2\pi -\frac{v_2}{r_2}\cdot t\\ \color{blue} \alpha_1=\alpha_2 \)

 

\(\frac{v_1}{r_1}\cdot t=2\pi -\frac{v_2}{r_2}\cdot t\\ (\frac{v_1}{r_1}+\frac{v_2}{r_2})\cdot t=2\pi\\ \color{blue}t=2\pi/(\frac{v_1}{r_1}+\frac{v_2}{r_2}) \)

 

\(t=2\pi/(\frac{250m}{min\cdot 50m}+\frac{300m}{min\cdot 60m})=2\pi \times 0,1min\ \times \frac{60sec}{min}=\color{blue}37.7sec\)

 

\(30\times 60sec /37.7sec=\color{blue}47.7\)

 

Odell and Kershaw meet 47 times.

laugh  !

 Nov 2, 2019

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