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The formulas above are used to compute the distance a car

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Eternal Intern
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The formulas above are used to compute the distance a car [#permalink] New post 26 Jul 2003, 14:23
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The formulas above are used to compute the distance a car travels after the brakes are applied. If the driver of a car applied the brakes just as a traffic light turned yellow and stopped exactly 6 seconds later, what is the value of V?
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 [#permalink] New post 28 Jul 2003, 10:10
Please explain the answer and how you got 20. I couldnt solve the sum at all.
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 [#permalink] New post 28 Jul 2003, 10:25
The car stopped exactly 6 seconds later (after applying the brakes)

So, 6 = 2d/v [from equation (2), t = 2d / v]
=> 2d = 6v
=> d = 3v

Putting that in equation 1, we get,

3v = 3v^2 / 20 [because d = 3v^2/20 from equation (1)]
=> v = v^2 / 20
=> 20v = v^2
=> v^2 – 20v = 0
=> v(v-20) = 0

So v=0 or v=20

Velocity cannot be 0, because that would imply that the car was not moving. So, v = 20

"If the driver of a car applied the brakes just as a traffic light turned yellow" is immaterial. Because if the guy was driving at 20mph and the above equations hold true, he will stop after 6 seconds. Where he applied the brakes(after light turned yellow, red etc.) will not have an effect on the distance traveled.
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 [#permalink] New post 28 Jul 2003, 11:27
You beat me to the punch:) :-D I was going to show Joe the soltution 8-)
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 [#permalink] New post 23 Jan 2004, 07:31
a quadratic equation ?

1) d = 3 * v^2 /20
2) t = 2*d/v

so from 2) d = t*v/2

now t * v/2 = 3 * v^2/20
so v = 10 * t /3
if t = 6 then v = 20
  [#permalink] 23 Jan 2004, 07:31
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The formulas above are used to compute the distance a car

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