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OpenStudy (anonymous):
a boat moves from p to q (downstream) in 2.5 h and from q to p (upstream) in 5 h respectively when the engie is producing maximum speed on the boat in each case . if due to increased water flow the speed of flow of the river gets doubled how long will it take the boat(moving with possible speed) to move from
(a) p to q (downstream) (b) q to p (upstream)
14 years ago
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OpenStudy (anonymous):
the answer is (a) 2h (b) 10h can anybody give me the steps plz...
14 years ago
OpenStudy (anonymous):
Downstream
\[L=Vr.t=(Vmax+vriver).t\]
Upstream
\[L=V'r.t'=(Vmax-vriver).t'\]
14 years ago
OpenStudy (anonymous):
ok
14 years ago
OpenStudy (anonymous):
solve it and then put v river times two
14 years ago
OpenStudy (anonymous):
what????
14 years ago
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OpenStudy (anonymous):
he said the vriver has doubled
14 years ago
OpenStudy (anonymous):
can u give the eq plzz
14 years ago
OpenStudy (anonymous):
ok wait
14 years ago
OpenStudy (anonymous):
vmax = 3s/10
14 years ago
OpenStudy (anonymous):
is that answer correct...
14 years ago
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OpenStudy (anonymous):
\[vmax=3.vriver\]
14 years ago
OpenStudy (anonymous):
oh....no
14 years ago
OpenStudy (anonymous):
after finding v max should we find speer of river???
14 years ago
OpenStudy (anonymous):
\[L=(Vmax+vriver).2,5=(3.vriver+vriver).2.5=4.vriver.2.5\]
14 years ago
OpenStudy (anonymous):
Now we doubled v river
\[L=(Vmax+2.vriver).t=(3.vriver+2vriver).t=4.vriver.2.5\]
so t= 2h
14 years ago
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OpenStudy (anonymous):
no idea !!
14 years ago
OpenStudy (anonymous):
what you dont get?
14 years ago
OpenStudy (anonymous):
L=(Vmax+2.vriver).t=(3.vriver+2vriver).t=4.vriver.2.5 ???????????
14 years ago
OpenStudy (anonymous):
here we find L as a function of vriver
\[L=(Vmax+vriver).2,5=(3.vriver+vriver).2.5=4.vriver.2.5\]
14 years ago
OpenStudy (anonymous):
plzz help ????
14 years ago
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OpenStudy (anonymous):
i 'll take a picture for you
14 years ago
OpenStudy (anonymous):
ok
14 years ago
OpenStudy (anonymous):
14 years ago
OpenStudy (anonymous):
14 years ago
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