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Speed-distance Algebra 1 Problem

  1. Nov 25, 2012 #1
    Toward the end of Middle School, I always knew it would come time for me to confront my algebraic innumeracy by means of rigorous and meticulous study of textbooks made available to me by friends who have taken Algebra 1 and Algebra 2 courses, to acquaint myself with what I need to know in order to take a set of courses that are math-intensive. I ask that you squelch whatever thoughts you have about me, a Grade 11 student who should be able to solve such problems without worry.

    Below is the question that stultifies me, accompanied by my uneasy undertaking to solve it.

    1. The problem statement, all variables and given/known data
    2gtw0sl.png
    2. Relevant equations
    s = d/t, I imagine.
    3. Attempt at a solution

    a. I'm confident with my response to this one:

    47x, 'x' being the number of hours the gazelle spent travelling: the totality of this expression upon the value of 'x' being discerned would represent distance travelled.

    b. I'm confident with my response to this one, too:

    65x, 'x' being the number of hours the cheetah spent travelling.

    c. Here's where it gets tricky (for me). What I got was:

    65x - 1

    My reasoning, simple: it takes x hours for the cheetah to catch the gazelle. I don't exactly know how the -1 comes into play here (and not +1, considering he has to run another mile).

    d. 65x = 47x + 1
    [Subtract 47x from both sides]
    18x = 1
    x = 1/18

    It would take 1/18 of an hour for the cheetah to catch up to the gazelle.

    If we take what I initially suspected the answer to 'c' to be, we get a negative yielding, which is totally implausible:

    65x + 1 = 47x
    65x = 57x - 1
    18x = -1
    x = -1/18

    Am I even on the right track?
     
  2. jcsd
  3. Nov 25, 2012 #2
    "My reasoning, simple: it takes x hours for the cheetah to catch the gazelle. I don't exactly know how the -1 comes into play here (and not +1, considering he has to run another mile)."

    This is wrong. The cheetah travels 65x miles in time x, while the gazelle travels 47x. As the cheetah approaches the gazelle, the gazelle will be going forward as well. This means that the cheetah won't run for 1/65 hours as you would expect. It will run more. Can you think how much it would run?
     
  4. Nov 25, 2012 #3

    HallsofIvy

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    If the gazelle runs at 47 mi/hr and the cheetah runs at 65 mi/h, how fast is the distace between them closing?
     
  5. Nov 25, 2012 #4
    Millennial: I fully understand the gazelle will be running too. If I didn't, I would have put for "c." that the cheetah would have to travel only one mile in order to catch up to the gazelle. lol

    Here is my second attempt. (HallsofIvy omits that there is a 1 mile disparity, and doesn't address my unease with representing this algebraically.)

    The distance between them closes when the difference between the respective distances travelled is 0:

    65*x - 47*x + 1 = 0
    18*x + 1 = 0
    18*x = -1
    x = -1/18

    In -1/18 of an hour, the cheetah will catch up to the gazelle.

    Hardly a plausible answer.

    @RV: I wasn't complaining. It was merely a statement of fact.
     
    Last edited: Nov 25, 2012
  6. Nov 25, 2012 #5

    symbolipoint

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    That is the right approach. The rate of approach is 65-47 miles per hour. The distance between them is decreasing by 18 miles per hour. How far from each other are they at the start of part c. ? How much time would be required to close the distance to 0? Now, how far was this for the cheetah, using the actual fastest speed for the cheetah?
     
  7. Nov 25, 2012 #6

    Ray Vickson

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    HallsofIvy did not omit anything; he simply left it up to you to do some of the work (as per Forum rules). Also: people cannot tell what makes you feel uneasy if you don't tell them!

    RGV
     
  8. Nov 25, 2012 #7

    symbolipoint

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    Clue: The TIME is the same but the distances are different.
     
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