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    lewit's Avatar
    lewit Posts: 2, Reputation: 1
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    #1

    Jul 13, 2012, 11:32 PM
    Quadratic equations given time and distance
    Dieter makes a journey of 430 km, travelling 160km by bus and 270km by car. If the car averages 10km/h faster than the bus and the entire journey takes 5h, what was the speed of the car?
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #2

    Jul 14, 2012, 04:31 AM
    Start with the idea that the time t required to cover some given distance d, when traveling at a rate r (where r is expressed in the same units as t and d; e.g. if t is in hours and d is in miles, then r is mph), is given by



    From the problem, let b denote the bus's speed and b + 10 denote the car's average speed.

    Thus your problem sets up as



    From here on it's just algebra, leading you to a quadratic expression to solve for b. (Hint: in this case you'll have two solutions, but disregard the negative one.)
    Cvholliday's Avatar
    Cvholliday Posts: 8, Reputation: 1
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    #3

    Jul 14, 2012, 04:38 AM
    Wow... this is a tough one. You can set up some equations using distance = rate x time. For example, for the car, let's call C the rate of the car and T the time Dieter is in the car. So we can write the equation CT=270. For the bus, it's rate must be C-10 since it is 10km/h than the car and the time on the bus is 5-T since the entire time of the trip is 5 hours. Now we can write, for the bus, (C-10)(5-T)=160 which we can solve with a quadratic only after we re-write T using C as a variable. Since CT=270, then T=270/C. So let's re-write our equation as (C-10)(5-270/C)=160. Solve that and you have C, the rate of the car.
    lewit's Avatar
    lewit Posts: 2, Reputation: 1
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    #4

    Jul 14, 2012, 07:08 AM
    Thank you for your help it makes a lot more sense now (:
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