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Math Help - Word Problems? I'm Lost

  1. #1
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    Exclamation Word Problems? I'm Lost

    I think I can do it once I get an equation, but I'm having trouble coming up with an equation.

    A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.

    How far has the car moved when its speed has been reduced to 30 miles per hour?
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  2. #2
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by sammy07 View Post
    I think I can do it once I get an equation, but I'm having trouble coming up with an equation.

    A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.

    How far has the car moved when its speed has been reduced to 30 miles per hour?
    First, convert to the same units:

    45mh^{-1} = 45(5280) fh^{-1} = 237600 fh^{-1} (feet per hour)

    It stops in \frac{132}{237600} = \frac{1}{1800} hours.

    Deceleration is \frac{|0-45|}{(\frac{1}{1800})}=\frac{1}{40}fh^{-2}<br />

    So you want to find \frac{1}{40}x = 30(5280) = 158400

    Solving gives x = 63360000 feet

    = 1200 miles.

    On second thought, I think it would have been better to have done the entire problem in terms of miles.
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  3. #3
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    Quote Originally Posted by DivideBy0 View Post
    First, convert to the same units:

    45mh^{-1} = 45(5280) fh^{-1} = 237600 fh^{-1} (feet per hour)

    It stops in \frac{132}{237600} = \frac{1}{1800} hours.

    Deceleration is \frac{|0-45|}{(\frac{1}{1800})}=\frac{1}{40}fh^{-2}<br />

    So you want to find \frac{1}{40}x = 30(5280) = 158400

    Solving gives x = 63360000 feet

    = 1200 miles.

    On second thought, I think it would have been better to have done the entire problem in terms of miles.
    That doesn't make sense. If I'm stopping witin 135 feet, I could not have gone 1200 miles.
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  4. #4
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    Problem with that solution is not only the answer but the fact that 1/1800 hours is not the time it takes to go from 45 to 0 mph. That is the time it takes to go 132 feet at 45 mph.

    The nature of this problem sounds a lot easier than it is. You may have to use proportions:

    \frac{132ft}{45mph} = \frac{x}{30mph}
    x = 88 feet
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  5. #5
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by sammy07 View Post
    That doesn't make sense. If I'm stopping witin 135 feet, I could not have gone 1200 miles.
    Sorry, I shouldn't be doing problems so late
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