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Rabbits Average Speed?

  1. Jun 23, 2007 #1
    Rabbits!!!! Average Speed??

    1. The problem statement, all variables and given/known data

    A rabbit takes part in a 4.00km race. First the rabbit runs 0.500 km then stops and takes a 90.0min nap. when he wakes up he runs twice as fast completing the race in a total time of 1.75h.

    A:-Calculate the average speed of the rabbit?
    B:-Calculate the average speed of the rabbit before he stopped for a nap??

    2. Relevant equations

    Average Speed = Total Distance Traveled / Total Time Taken

    3. The attempt at a solution

    I got the 1st one, A, the answer to that was 2.29km/h......

    Ok i have been trying to sole the 2nd part of this myth but i still cant figure out how to get the 9.00 km/hr.. my attempt is below

    Converting 90 min to hrs = 1.5h...therefore actual time taken for the 2 distances would = to 0.25h ie 1.75h - 1.5h

    Thus forming the 1st equation T1 + T2 = 0.25h --------(1)

    v = avg. speed; d = distance; t = time

    now we know that v1 = d1 / t1 and v2 = d2 / t2......
    where d1 = 0.5km and d2 = 3.5km
    and according to the question we can form another equation stating that

    2*v1 = v2 -----------------(2)

    using the given formula we can write an equation like

    (d1 / v1) + (d2 / v2) = t1 + t2

    (0.5km / v1) + (3.5km / 2*v1) = 0.25h

    (4.0km / 3*v1) = 0.25h

    3*0.25h*v1 = 4.0km

    therefore v1 = 4.0km / 0.75h = 5.34 km/hr...........

    WHAT AM DOING WRONG?????????

    according to my book the answer is 9.00km/h :confused:

    Please Help........
     
    Last edited: Jun 23, 2007
  2. jcsd
  3. Jun 24, 2007 #2

    danago

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    Gold Member

    Check your working in solving the equation. You have made an error in adding the fractions.

    Note that:
    [tex]
    \frac{a}{b} + \frac{x}{y} \ne \frac{{a + x}}{{b + y}}
    [/tex]
     
  4. Jun 24, 2007 #3
    But according to the question the rabbit runs twice as fast so isn't it correct to say that 2 times V1 would equal to V2 !!! therefore substituting it in the problem..where v2 = 2 * v1...
     
  5. Jun 24, 2007 #4

    danago

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    Gold Member

    yes that part is right, so this is correct:

    (0.5km / v1) + (3.5km / 2*v1) = 0.25h

    But remeber, when you add fractions, you need to find a common denominator.
     
  6. Jun 24, 2007 #5
    ok then what is wrong with my final answer.....why is the book saying its 9.00km/hr
     
  7. Jun 24, 2007 #6

    danago

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    Gold Member

    You didnt add the fractions correctly.

    (0.5km / v1) + (3.5km / 2*v1) [tex]\ne[/tex] (4.0km / 3*v1)

    But rather:

    (0.5km / v1) + (3.5km / 2*v1) [tex]=[/tex] (4.5km / 2*v1)
     
  8. Jun 24, 2007 #7
    Darn what a silly mistake >_<! Thanks a million..........came out right!!!
     
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