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Thread: Two Cars

  1. #1
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    Two Cars

    Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives st Wildwood at 11:00 am, 1/2 hour before the other car. What was the average speed of each car? How far did each travel?

    Car 1: rate = x + 10; time = 3 hours

    Car 2: rate = x; time = 3.5 hours

    Equation to find the average speed of each car is:

    3/(x + 10) = x/(3.5)

    After finding x, I can use D = rt TWICE to find how far each car travelled.

    Correct?
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    Re: Two Cars

    Quote Originally Posted by harpazo View Post
    Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives st Wildwood at 11:00 am, 1/2 hour before the other car. What was the average speed of each car? How far did each travel?

    Car 1: rate = x + 10; time = 3 hours

    Car 2: rate = x; time = 3.5 hours

    Equation to find the average speed of each car is:

    3/(x + 10) = x/(3.5)
    x is a rate with units of, say, miles/hour while "3" and "3.5" are times, in hours. The fraction on the left has units of hours/(miles per hour)= hours^2/miles while the fraction on the right has units of (miles/hour)/hours= miles/hour^2. They can't be equal!

    After finding x, I can use D = rt TWICE to find how far each car travelled.

    Correct?
    No, your equation makes no sense. If the first car has speed x+ 10 and the distance from Commercial Boulevard to Wildwood is "d" then the time taken is d/(x+ 10). The time taken by the second car, at speed x is d/x. Saying that the first car arrived at 11:00 means that d/(x+ 10)= 3. Saying the first car arrived 1/2 an hour ahead of the second means that d/x= 3.6. Solve those equations for x and d.

    Exactly how are you interpreting " How far did each travel?" In what time span? The "d" above is the distance the first car travels in 3 hours. Are you interpreting this as asking how far the second car traveled in that same time?
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    Re: Two Cars

    Next time you get one of these, DRAW A DIAGRAM (as is, you seem to be guessing!)

    (8.00am)@x.............3.5x.............>11.30am

    (8.00am)@x+10.......3(x+10)........>11.00am

    3(x+10) = 3.5x
    Last edited by DenisB; Nov 7th 2018 at 09:06 AM.
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    Re: Two Cars

    Quote Originally Posted by HallsofIvy View Post
    x is a rate with units of, say, miles/hour while "3" and "3.5" are times, in hours. The fraction on the left has units of hours/(miles per hour)= hours^2/miles while the fraction on the right has units of (miles/hour)/hours= miles/hour^2. They can't be equal!


    No, your equation makes no sense. If the first car has speed x+ 10 and the distance from Commercial Boulevard to Wildwood is "d" then the time taken is d/(x+ 10). The time taken by the second car, at speed x is d/x. Saying that the first car arrived at 11:00 means that d/(x+ 10)= 3. Saying the first car arrived 1/2 an hour ahead of the second means that d/x= 3.6. Solve those equations for x and d.

    Exactly how are you interpreting " How far did each travel?" In what time span? The "d" above is the distance the first car travels in 3 hours. Are you interpreting this as asking how far the second car traveled in that same time?
    Quote Originally Posted by DenisB View Post
    Next time you get one of these, DRAW A DIAGRAM (as is, you seem to be guessing!)

    (8.00am)@x.............3.5x.............>11.30am

    (8.00am)@x+10.......3(x+10)........>11.00am

    3(x+10) = 3.5x
    I am selecting questions from the textbook that the author did not give an example for. This is what matters to me. Selection of questions that the author has fully explained in the chapter example section is too easy. My goal is to rationalize my way through word problems not given as examples at the beginning of each chapter.
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    Re: Two Cars

    Quote Originally Posted by DenisB View Post
    Next time you get one of these, DRAW A DIAGRAM (as is, you seem to be guessing!)

    (8.00am)@x.............3.5x.............>11.30am

    (8.00am)@x+10.......3(x+10)........>11.00am

    3(x+10) = 3.5x
    No, I am not guessing. Read my reply to
    HallsofIvy.
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    Re: Two Cars

    3(x+10) = 3.5x

    3x + 30 = 3.5x

    3x - 3.5x = -30

    -0.5x = -30

    x = -30/-0.5

    x = 60

    Distance travelled by car 1:

    d = 3x + 30

    d = 3(60) + 30

    d = 180 + 30

    d = 210 miles

    Distance travelled by car 2:

    d = 3.5x

    d = 3.5(60)

    d = 210 miles

    Average speed for car 1:

    speed = distance/time

    speed = 210/3

    Speed = 70 mph

    Average speed for car 2:

    Speed = distance/time

    Speed = 210/3.5

    Speed = 60 mph
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    Re: Two Cars

    For these sorts of problems, ie when you have three variables (eg speed, distance, time) and know a relationship between them (eg s=d/t), I always like to set up a table such as:

    Two Cars-capture.jpg

    and develop your equation from there.

    Since Distance = Speed x time, we get d =3.5x for Car1 and d=3(x+10) for Car2.

    Therefore:


    3.5x = 3(x+10) and go from there.
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    Re: Two Cars

    Quote Originally Posted by harpazo View Post
    Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives st Wildwood at 11:00 am, 1/2 hour before the other car. What was the average speed of each car? How far did each travel?
    Car 1: rate = x + 10; time = 3 hours
    Car 2: rate = x; time = 3.5 hours
    Equation to find the average speed of each car is:
    3/(x + 10) = x/(3.5)
    Next time you try something similar, equate the DISTANCE instead:
    3(x + 10) = 3.5x

    Much easier...
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    Re: Two Cars

    Quote Originally Posted by Debsta View Post
    For these sorts of problems, ie when you have three variables (eg speed, distance, time) and know a relationship between them (eg s=d/t), I always like to set up a table such as:

    Click image for larger version. 

Name:	Capture.JPG 
Views:	5 
Size:	19.9 KB 
ID:	39111

    and develop your equation from there.

    Since Distance = Speed x time, we get d =3.5x for Car1 and d=3(x+10) for Car2.

    Therefore:


    3.5x = 3(x+10) and go from there.
    Quote Originally Posted by DenisB View Post
    Next time you try something similar, equate the DISTANCE instead:
    3(x + 10) = 3.5x

    Much easier...
    See my full reply above. Is it right?
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    Re: Two Cars

    Quote Originally Posted by harpazo View Post
    See my full reply above. Is it right?
    Ya. But way too long...
    On a timed test, you'd be in trouble!

    We're trying to lend you an electric shovel to dig a hole,
    but you keep using a kitchen fork
    Last edited by DenisB; Nov 9th 2018 at 10:22 AM.
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    Re: Two Cars

    Quote Originally Posted by DenisB View Post
    Ya. But way too long...
    On a timed test, you'd be in trouble!

    We're trying to lend you an electric shovel to dig a hole,
    but you keep using a kitchen fork
    DenisB:

    I am 53 years old. What timed test are you talking about? I ended my school days in the Fall semester 1993.
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    Re: Two Cars

    Ya...but if you do start tutoring, you'll need to know the "quicker ways"
    since students are given "timed tests"...like 1/2 hour for 6 problems.

    Your age 53 is young...I'm 77 !
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    Re: Two Cars

    See my full reply above. Is it right?

    Yes it is!
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  14. #14
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    Re: Two Cars

    Quote Originally Posted by DenisB View Post
    Ya...but if you do start tutoring, you'll need to know the "quicker ways"
    since students are given "timed tests"...like 1/2 hour for 6 problems.

    Your age 53 is young...I'm 77 !
    I had no idea that you are in your 70s. You are young too compared to other members. I recall Soroban telling me back in 2006 that he was in his 80s. BTW, what happened to MHF member Soroban? He was my online tutor at the time.
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    Re: Two Cars

    YES: I remember Soroban well; inactive for a few years now. Quite the guy!!
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