Question-50093

Question Number 50093 by peter frank last updated on 13/Dec/18

Commented by peter frank last updated on 13/Dec/18

Answered by peter frank last updated on 14/Dec/18

momentum=mv r^′ =v=(8t−3t^2 )i+−5j+4t^3 k r^(′′) =a=(8−6t)i+12t^2 k momentum=0.002[(8t−3t^2 )i+−5j+4t^3 k]kgms^(-1) ii)F=ma t=1sec F=0.002[(8−6t)i+12t^2 k] F=0.002[(8−6(1))i+12(1^2 )k] F=0.002[(2)i+12k]

$$\mathrm{momentu}{m}={mv} \\ $$$${r}^{'} ={v}=\left(\mathrm{8t}−\mathrm{3t}^{\mathrm{2}} \right){i}+−\mathrm{5}{j}+\mathrm{4}{t}^{\mathrm{3}} {k} \\ $$$${r}^{''} ={a}=\left(\mathrm{8}−\mathrm{6t}\right){i}+\mathrm{12}{t}^{\mathrm{2}} {k} \\ $$$${momentum}=\mathrm{0}.\mathrm{002}\left[\left(\mathrm{8t}−\mathrm{3t}^{\mathrm{2}} \right){i}+−\mathrm{5}{j}+\mathrm{4}{t}^{\mathrm{3}} {k}\right]{kgms}^{-\mathrm{1}} \\ $$$$\left.\mathrm{ii}\right){F}=\mathrm{ma}\:\:\:\:{t}=\mathrm{1}{sec} \\ $$$${F}=\mathrm{0}.\mathrm{002}\left[\left(\mathrm{8}−\mathrm{6t}\right){i}+\mathrm{12}{t}^{\mathrm{2}} {k}\right] \\ $$$${F}=\mathrm{0}.\mathrm{002}\left[\left(\mathrm{8}−\mathrm{6}\left(\mathrm{1}\right)\right){i}+\mathrm{12}\left(\mathrm{1}^{\mathrm{2}} \right){k}\right] \\ $$$${F}=\mathrm{0}.\mathrm{002}\left[\left(\mathrm{2}\right){i}+\mathrm{12}{k}\right] \\ $$

Answered by peter frank last updated on 13/Dec/18

mkv^3 =−F mkv^3 =−ma=−m(dv/dt) kv^3 =−(dv/dx).(dx/dt) kv^3 =−v(dv/dt) −(dv/v^2 )=kdx ∫_u ^v −(dv/v^2 )=∫_0 ^x kdx (1/v)−(1/u)=kx V=(u/(kUx+1)) ii) V=(dx/dt)=(u/(kUx+1)) ∫_0 ^x (kUx+1)dx=∫_0 ^1 udt t=(((1/2)kUx^2 +x)/U) t=(x/u)+(1/2)kx^2

$${mkv}^{\mathrm{3}} =−{F} \\ $$$${mkv}^{\mathrm{3}} =−{ma}=−{m}\frac{{dv}}{{dt}} \\ $$$${kv}^{\mathrm{3}} =−\frac{{dv}}{{dx}}.\frac{{dx}}{{dt}} \\ $$$${kv}^{\mathrm{3}} =−{v}\frac{{dv}}{{dt}} \\ $$$$−\frac{{dv}}{{v}^{\mathrm{2}} }={kdx} \\ $$$$\int_{\mathrm{u}} ^{\mathrm{v}} −\frac{{dv}}{{v}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{x}} {kdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{v}}−\frac{\mathrm{1}}{\mathrm{u}}={kx} \\ $$$${V}=\frac{{u}}{{kUx}+\mathrm{1}} \\ $$$$\left.{ii}\right) \\ $$$${V}=\frac{{dx}}{{dt}}=\frac{{u}}{{kU}\mathrm{x}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{{x}} \left({kU}\mathrm{x}+\mathrm{1}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{udt} \\ $$$${t}=\frac{\frac{\mathrm{1}}{\mathrm{2}}{kUx}^{\mathrm{2}} +{x}}{{U}} \\ $$$${t}=\frac{{x}}{{u}}+\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} \\ $$$$ \\ $$

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