એક બોટ નદીના બે સ્થળો વચ્ચેના અમુક અંતરને આવર | vedclass

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Question

(c)

For upstream, Speed $\Rightarrow v-u$

(where $v \rightarrow$ man and $u \rightarrow$ water)

For downstream, Speed $\Rightarrow v+u$

$t

Vedclass · Accepted Answer

$\frac{2 t_1 t_2}{t_1+t_2}$

Vedclass · Answer

(c)

For upstream, Speed $\Rightarrow v-u$

(where $v ightarrow$ man and $u ightarrow$ water)

For downstream, Speed $\Rightarrow v+u$

$t_{ up }=\frac{d}{v-u}$

$t_2=\frac{d}{v-u}$

$\Rightarrow d=(v-u) t_2 \ldots (i)$

$t_{ up }=\frac{d}{v-u}$

$t_2=\frac{d}{v-u}$

$\Rightarrow d=(v-u) t_2 \ldots (ii)$

$t_{ still }=\frac{d}{v}$

$t_{	ext {still }}=\frac{2 t_1 t_2}{t_1+t_2}$

On equating $(i)$ and $(ii)$

$(v-u) t_2=(v+u) t_1$

$\Rightarrow v t_2-u t_2=v t_1+u t_1$

$\Rightarrow v\left(t_2-t_1ight)=u\left(t_1+t_2ight)$

$\Rightarrow u=\frac{v\left(t_2-t_1ight)}{t_2+t_1}$

So, $d=\left(v-\frac{v\left(t_2-t_1ight)}{t_1+t_2}ight) t_2=v t_2\left(\frac{t_1+t_2-t_2+t_1}{t_1+t_2}ight)$

$\frac{d}{v}=\frac{2 t_1 t_2}{t_1+t_2} ightarrow$ Remember as shortcut

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