Question-225397

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Question Number 225397 by fantastic last updated on 24/Oct/25

Commented by fantastic last updated on 24/Oct/25

$${looks}\:{easy}\:{right}? \\ $$

Answered by mr W last updated on 25/Oct/25

Commented by fantastic last updated on 25/Oct/25

$${i}\:{think}\:{i}\:{have}\:{to}\: \\ $$$${practice}\:{alot}\:{to}\:{build} \\ $$$${an}\:{approach}\:{like}\:{you}.\: \\ $$$${TY}\:{sir} \\ $$

Commented by mr W last updated on 25/Oct/25

a=(dv/dt)=v(dv/dx)=−kv^2 with k=constant (dv/dx)=−kv (dv/v)=−kdx ∫_v_0 ^v_1 (dv/v)=−k∫_0 ^h dx ln v_1 −ln v_0 =−kh ⇒k=((ln v_0 −ln v_1 )/h) (dv/dt)=−(((ln v_0 −ln v_1 )/h))v^2 ∫_v_0 ^v_1 (dv/v^2 )=−(((ln v_0 −ln v_1 )/h))∫_0 ^T dt −((1/v_1 )−(1/v_0 ))=−(((ln v_0 −ln v_1 )/h))T ⇒T=(((v_0 −v_1 )h)/(v_0 v_1 (ln v_0 −ln v_1 ))) ✓

$${a}=\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dx}}=−{kv}^{\mathrm{2}} \:{with}\:{k}={constant} \\ $$$$\frac{{dv}}{{dx}}=−{kv} \\ $$$$\frac{{dv}}{{v}}=−{kdx} \\ $$$$\int_{{v}_{\mathrm{0}} } ^{{v}_{\mathrm{1}} } \frac{{dv}}{{v}}=−{k}\int_{\mathrm{0}} ^{{h}} {dx} \\ $$$$\mathrm{ln}\:{v}_{\mathrm{1}} −\mathrm{ln}\:{v}_{\mathrm{0}} =−{kh} \\ $$$$\Rightarrow{k}=\frac{\mathrm{ln}\:{v}_{\mathrm{0}} −\mathrm{ln}\:{v}_{\mathrm{1}} }{{h}} \\ $$$$\frac{{dv}}{{dt}}=−\left(\frac{\mathrm{ln}\:{v}_{\mathrm{0}} −\mathrm{ln}\:{v}_{\mathrm{1}} }{{h}}\right){v}^{\mathrm{2}} \\ $$$$\int_{{v}_{\mathrm{0}} } ^{{v}_{\mathrm{1}} } \frac{{dv}}{{v}^{\mathrm{2}} }=−\left(\frac{\mathrm{ln}\:{v}_{\mathrm{0}} −\mathrm{ln}\:{v}_{\mathrm{1}} }{{h}}\right)\int_{\mathrm{0}} ^{{T}} {dt} \\ $$$$−\left(\frac{\mathrm{1}}{{v}_{\mathrm{1}} }−\frac{\mathrm{1}}{{v}_{\mathrm{0}} }\right)=−\left(\frac{\mathrm{ln}\:{v}_{\mathrm{0}} −\mathrm{ln}\:{v}_{\mathrm{1}} }{{h}}\right){T} \\ $$$$\Rightarrow{T}=\frac{\left({v}_{\mathrm{0}} −{v}_{\mathrm{1}} \right){h}}{{v}_{\mathrm{0}} {v}_{\mathrm{1}} \left(\mathrm{ln}\:{v}_{\mathrm{0}} −\mathrm{ln}\:{v}_{\mathrm{1}} \right)}\:\:\:\checkmark \\ $$

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