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Can you do it using variable and s=d/t formula

Module 0 Day 9 Your Turn Part 3
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  • F
    fabulousgrizzly M0
    last edited by debbie Jun 19, 2020, 3:02 AM Jun 17, 2020, 7:39 PM

    Module 0 Week 3 Day 9 Your Turn Explanation Part 3

    Can you please explain to me how Prof Loh did the ratio problem. I didn't understand it at all? If possible can you do it using variable and s=d/t formula because I am a physics guy and is used to variables and so understand the concept better by looking at variables. Thank you!

    D 1 Reply Last reply Jun 18, 2020, 12:49 AM Reply Quote 1
    • D
      debbie ADMIN M0★ M1 M5 @fabulousgrizzly
      last edited by debbie Jun 18, 2020, 1:26 AM Jun 18, 2020, 12:49 AM

      @fabulousgrizzly Certainly, I'm happy to help explain two different ways to solve this Your Turn question, which I'll repeat here for convenience: "You're driving a freight train away from Los Angeles at 20 miles per hour, and notice that passenger trains pass you from behind every 60 minutes, moving at 60 miles per hour. How many minutes apart do the passenger trains depart from Los Angeles?"

       


      First way (Rate equation):

      The first step is to find the distance between the faster passenger trains. Once we have this, then we can solve for the minutes between each passenger train departure from the station.

      M0W3D9-y-forum-explanation-just-passenger-trains-600.png

       

      This distance is equivalent to the extra distance that a passenger train overtakes us by after one hour.

      Why is this true? It's because after one hour, the second train has caught up to us, so the distance between passenger trains is really the distance between our train and first passenger train.

      M0W3D9-y-forum-explanation-simple.png

      How many miles did the faster passenger train overtake us by?

      Since our train goes at 202020 mph and the passenger train goes at 606060 mph, the relative speed of the train compared to us is

       relative speed =60−20=40 mileshour \text{ relative speed } = 60 - 20 = 40 \text{ } \frac{\text{miles}}{\text{hour}}  relative speed =60−20=40 hourmiles​

      We want the distance between the first passenger train and our train after 111 hour, so here t=1 hour,t = 1 \text{ } \text{hour},t=1 hour, and we can use the rate equation to solve for the distance d:d:d:

      d=vrelativetd=40 mileshour×1 hourd=40 miles\begin{aligned} d &= v_{relative}t \\ d &= 40 \text{ } \frac{\text{miles}}{\text{hour}} \times 1 \text{ } \text{hour} \\ d &= 40 \text{ } \text{miles} \\ \end{aligned} ddd​=vrelative​t=40 hourmiles​×1 hour=40 miles​

      Then, since the second passenger train has caught up to us by this time,

       distance between us and the first passenger train = distance between the first passenger train and the second passenger train \text{ distance between us and the first passenger train } = \text{ distance between the first passenger train and the second passenger train} distance between us and the first passenger train = distance between the first passenger train and the second passenger train

      Thus there are 40  miles 40 \text{ } \text{ miles }40  miles  between two adjacent passenger trains, and we can find the hours between their departures using

      t=dvt=40  miles60 mileshourt=23 hour\begin{aligned} t &= \frac{d}{v} \\ t &= \frac{40 \text{ } \text{ miles} }{60 \text{ } \frac{\text{miles}}{\text{hour}}} \\ t &= \frac{2}{3} \text{ } \text{hour} \\ \end{aligned} ttt​=vd​=60 hourmiles​40  miles​=32​ hour​

      Converting this to minutes, we get 23 hour×60minhour=40 minutes.\frac{2}{3} \text{ } \text{hour} \times 60 \frac{\text{min}}{\text{hour}} = \boxed{40 \text{ } \text{minutes.}} 32​ hour×60hourmin​=40 minutes.​


      Second way (Prof. Loh's Way, Using a diagram):

      M0W3D9-y-forum-explanation-our-train.png

       
      Our train goes 202020 miles in one hour. Now let's add in the passenger train, which has caught up to us at the start, where t=0 minutes.t = 0 \text{ } \text{minutes}.t=0 minutes.
       

      M0W3D9-y-forum-explanation-diagram-0-min.png

      After 111 hour, the passenger train has covered 606060 miles, which is three times the distance that we have gone. Let's mark off three sections: we travel one section, and the passenger train travels three sections.

       
      M0W3D9-y-forum-explanation-diagram-60-min.png
       

      The next passenger train, colored in purple, \textcolor{purple}{\text{purple}},purple, is there too.

      Now, to find the distance between the two passenger trains, we just count the number of sections. There are two sections, and each section corresponded to 202020 miles, so there are 404040 miles between adjacent passenger trains.

      To find the time between departures of the passenger trains, notice that they go at the speed of 60 mileshour,60 \text{ } \frac{\text{miles}}{\text{hour}},60 hourmiles​, which is the same as 1 milemin.1 \text{ } \frac{\text{mile}}{\text{min}}.1 minmile​.

      The passenger train takes 404040 minutes to travel 404040 miles, so the time between departures of the passenger train is 40 minutes.\boxed{40 \text{ } \text{minutes}}.40 minutes​.

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