Quadratic equations given time and distance

ArcSine · 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

$t = \frac{d}{r}$

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

$5 = \frac{160}{b} + \frac{270}{b + 10}$

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 · 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 · Jul 14, 2012, 07:08 AM

Thank you for your help it makes a lot more sense now (:

Quadratic equations given time and distance

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