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Question Number 896 by 112358 last updated on 15/Apr/15

Given that the velocity v of a body  t seconds after passing a point O  is found by          v^2 =(1/k)[P−(P−kv_o ^2 )e^(−((2kt)/m)) ]  determine the distance covered  by the body  one hour after   passing O. The body moves along  a straight line. k,P,v_o  and m  are constants and t is measured in  seconds.

$${Given}\:{that}\:{the}\:{velocity}\:{v}\:{of}\:{a}\:{body} \\ $$$${t}\:{seconds}\:{after}\:{passing}\:{a}\:{point}\:{O} \\ $$$${is}\:{found}\:{by} \\ $$$$\:\:\:\:\:\:\:\:{v}^{\mathrm{2}} =\frac{\mathrm{1}}{{k}}\left[{P}−\left({P}−{kv}_{{o}} ^{\mathrm{2}} \right){e}^{−\frac{\mathrm{2}{kt}}{{m}}} \right] \\ $$$${determine}\:{the}\:{distance}\:{covered} \\ $$$${by}\:{the}\:{body}\:\:{one}\:{hour}\:{after}\: \\ $$$${passing}\:{O}.\:{The}\:{body}\:{moves}\:{along} \\ $$$${a}\:{straight}\:{line}.\:{k},{P},{v}_{{o}} \:{and}\:{m} \\ $$$${are}\:{constants}\:{and}\:{t}\:{is}\:{measured}\:{in} \\ $$$${seconds}. \\ $$$$ \\ $$

Commented by 123456 last updated on 15/Apr/15

∫_0 ^(3600) (√((1/k)[P−(P−kv_0 ^2 )e^(−((2kt)/m)) ]))dt  ∫_0 ^(3600) (√((P/k)−((P−kv_0 ^2 )/k)e^(−((2kt)/m)) ))dt  ∫_0 ^(3600) (√(α+βe^(γt) ))dt  α=(P/k)  β=((kv_0 ^2 −P)/k)  γ=−((2k)/m)

$$\underset{\mathrm{0}} {\overset{\mathrm{3600}} {\int}}\sqrt{\frac{\mathrm{1}}{{k}}\left[{P}−\left({P}−{kv}_{\mathrm{0}} ^{\mathrm{2}} \right){e}^{−\frac{\mathrm{2}{kt}}{{m}}} \right]}{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{3600}} {\int}}\sqrt{\frac{{P}}{{k}}−\frac{{P}−{kv}_{\mathrm{0}} ^{\mathrm{2}} }{{k}}{e}^{−\frac{\mathrm{2}{kt}}{{m}}} }{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{3600}} {\int}}\sqrt{\alpha+\beta{e}^{\gamma{t}} }{dt} \\ $$$$\alpha=\frac{{P}}{{k}} \\ $$$$\beta=\frac{{kv}_{\mathrm{0}} ^{\mathrm{2}} −{P}}{{k}} \\ $$$$\gamma=−\frac{\mathrm{2}{k}}{{m}} \\ $$

Commented by prakash jain last updated on 16/Apr/15

∫(√(α+βe^(γx) )) dx=((2((√(α+βe^(γx) ))−(√α)tanh^(−1) (((√(α+βe^(γx) ))/(√α)))))/γ)

$$\int\sqrt{\alpha+\beta{e}^{\gamma{x}} }\:{dx}=\frac{\mathrm{2}\left(\sqrt{\alpha+\beta{e}^{\gamma{x}} }−\sqrt{\alpha}\mathrm{tanh}^{−\mathrm{1}} \left(\frac{\sqrt{\alpha+\beta{e}^{\gamma{x}} }}{\sqrt{\alpha}}\right)\right)}{\gamma} \\ $$

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