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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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.
you are right
what do u think according to the statement of the question.......ist one or 2nd
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they reach B at the same time
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.....
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.
v1 = v*[1-(1/t)]
check this from back answers...........
is this right?
correct answer is (3V+2Vt)/2t
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....