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anonymous

  • one year ago

Two buses take the road to city B from city A. The speed of the slow one is V km/h. The fast one sets off 1 hour later than the slow one. It catches the slow bus in t hours at city B. Find the speed of the fast bus in terms of V and t.

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  1. anonymous
    • one year ago
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    see the faster bus must have caught the slower bus on the time when it just reaches the city B. do u think i am right ? or u think in this way that the slower bus has reached a long while ago in city B and the faster bus has reached there in t hrs.

  2. anonymous
    • one year ago
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    you are right

  3. anonymous
    • one year ago
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    what do u think according to the statement of the question.......ist one or 2nd

  4. anonymous
    • one year ago
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    they reach B at the same time

  5. anonymous
    • one year ago
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    according to the first statement... slower bus started 1 hr earlier than the faster one.... if faster one reached there in t time then the slower one got the (t-1) hrs. time to reach there.....

  6. anonymous
    • one year ago
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    if dist b/w A & B = x thus x/v =t-1.........................(1) speed of faster one =v1 thus x/t =v1............................(2) put the value of x from (1) in (2) to get the required eqn.

  7. anonymous
    • one year ago
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    v1 = v*[1-(1/t)]

  8. anonymous
    • one year ago
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    check this from back answers...........

  9. anonymous
    • one year ago
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    is this right?

  10. anonymous
    • one year ago
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    correct answer is (3V+2Vt)/2t

  11. anonymous
    • one year ago
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    see do u know the concept of relative velocity... this question is based on that.... i am not getting the correct eqn. but a MAJOR HINT the dist. covered by slower bus in 1 hr. time has to be covered up by the faster one which starts one hr. late.... so relative velocity of faster one wrt. to slower one is going to cover the relative displacement....

  12. anonymous
    • one year ago
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    relative displacement b/w the 2 buses...

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