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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

$$\mathrm{momentu}m=mv$$$$r^{{}^{\prime }}=v=(8\mathrm{t}-{3\mathrm{t}}^2)i+-5j+4t^{3}k$$$$r^{{}^{″}}=a=(8-6\mathrm{t})i+12t^{2}k$$$$momentum=0.002[(8\mathrm{t}-{3\mathrm{t}}^2)i+-5j+4t^{3}k]{kgms}^{-1}$$$$\mathrm{ii})F=\mathrm{ma}t=1sec$$$$F=0.002[(8-6\mathrm{t})i+12t^{2}k]$$$$F=0.002[(8-6\left(1\right))i+12\left(1^2\right)k]$$$$F=0.002[\left(2\right)i+12k]$$

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

$${mkv}^3=-F$$$${mkv}^3=-ma=-m\frac{dv}{dt}$$$${kv}^3=-\frac{dv}{dx}.\frac{dx}{dt}$$$${kv}^3=-v\frac{dv}{dt}$$$$-\frac{dv}{v^2}=kdx$$$${\int }_{\mathrm{u}}^{\mathrm{v}}-\frac{dv}{v^2}={\int }_0^{\mathrm{x}}kdx$$$$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=kx$$$$V=\frac{u}{kUx+1}$$$$ii)$$$$V=\frac{dx}{dt}=\frac{u}{kU\mathrm{x}+1}$$$${\int }_0^x(kU\mathrm{x}+1)\mathrm{dx}={\int }_0^1\mathrm{udt}$$$$t=\frac{\frac{1}{2}{kUx}^2+x}{U}$$$$t=\frac{x}{u}+\frac{1}{2}{kx}^2$$$$$$

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