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Distance and speed word problem. Introduction to vectors.

  1. Jan 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Two cars start from the same point at the same time, with one car traveling at a constant speed of 55 mph, and the other car having a constant speed of 70 mph. How far has the faster car travelled when it has a 15 minute lead on the slower car?

    Problems like this always throw me off.. I need help on the process of this problem.


    2. Relevant equations



    3. The attempt at a solution

    Not sure how to set it up really. I think I have to use D=r/t

    I also know the answer is in miles because it's distance, other than that I'm uncertain.
     
  2. jcsd
  3. Jan 31, 2013 #2

    cepheid

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    For motion at a constant speed:

    distance = speed*time (d = vt)

    You can use this to set up an expression for the distance vs. time of each car. Let car 1 be the v1 = 55 mph car and car 2 be the v2 = 70 mph car. At t = 0, d1 = d2 = 0 (they start at the same point).

    Now write down the equations for distance vs. time for each car. How far has each one travelled after t =15 minutes?
     
  4. Jan 31, 2013 #3
    So would that just be the velocity (70 and 55) times the time (15)?
    My answer choices are
    58 mi

    64 mi

    44 mi

    72 mi

    I'm thinking 44 with that information
     
  5. Jan 31, 2013 #4

    cepheid

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    I am guilty of misintepreting the word problem myself. For the faster car to have a 15 minute lead over the slower one, it means that the distance by which it is *gaining* over the slower car is such that it would take the slower car 15 min to make up for that lag (if it could...which it can't, because the lag grows, of course).


    EDIT: for clarity: enough time has elapsed that the faster car is 15 min ahead of the slow car, at slow car speed.

    So, first you have to figure how an equation for the *difference* between the distance travelled by the two cars vs. time. Then you have to set this difference equal to the distance that the slower car can travel in 15 minutes. Solve for the time it took for this distance (gain) to accumulate, and figure out how far the faster car travelled in that amount of time.
     
    Last edited: Jan 31, 2013
  6. Jan 31, 2013 #5
    I get the concept now just not how to get the equation setup
     
  7. Jan 31, 2013 #6

    cepheid

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    Did you write down this equation?

    This paragraph above basically walks you through it step by step. For the part in bold: you know an equation for d1 vs. time, and you know an equation for d2 vs time, so finding an equation for their difference is trivial.
     
  8. Feb 1, 2013 #7
    The only way I feel I can solve for distance is multiplying both speeds by 15 but if I subtract them I get 225. And how do I know how much the slower car can go in 15 minutes?
     
  9. Feb 1, 2013 #8

    cepheid

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    First of all, you shouldn't be getting anything nearly as large as 225. Are you remembering to include your UNITS? Your speeds are given in miles per HOUR. Your times are given in MINUTES. You need to have consistent units.

    Secondly, let me go through my hint again, with one small addition:

    If the distance travelled by the fast car is d1, and the distance travelled by the slow car is d2, then the difference between them (how far ahead the fast car is) is d1 - d2. You have an expression for d1 vs. time. You have an expression for d2 vs. time. Therefore, you have an expression for d1 - d2 vs. time. Come on. I can't be any more explicit without just doing the calculations for you.

    Really? You know its speed. You know the time interval (15 minutes). We already established above that distance = speed*time.
     
  10. Feb 1, 2013 #9
    So.... I think I got it.

    Since the slower car is 15 minutes behind it's 55mph * 15/60 = 13.75 miles.

    Which means that the faster car is 13.75 miles ahead

    so R1(t) = R2(t) + 13.75
    Or 70(t) - 55(t) = 13.75
    = 15t = 13.75
    = .917 hours
    D = r*t
    .917 * 70 = 64 miles (rounded)
     
  11. Feb 1, 2013 #10

    cepheid

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    Looks good to me!
     
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