A-particle-starts-from-the-origin-with-velocity-44-ms-1-on-a-straight-horizontal-road-Its-acceleration-varies-with-displacement-as-shown-The-velocity-of-the-particle-as-it-passes-through-th

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Question Number 16616 by Tinkutara last updated on 24/Jun/17

$$\mathrm{A}\mathrm{particle}\mathrm{starts}\mathrm{from}\mathrm{the}\mathrm{origin}\mathrm{with}$$$$\mathrm{velocity}\sqrt{44}{\mathrm{ms}}^{-1}\mathrm{on}\mathrm{a}\mathrm{straight}$$$$\mathrm{horizontal}\mathrm{road}.\mathrm{Its}\mathrm{acceleration}\mathrm{varies}$$$$\mathrm{with}\mathrm{displacement}\mathrm{as}\mathrm{shown}.\mathrm{The}$$$$\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{as}\mathrm{it}\mathrm{passes}$$$$\mathrm{through}\mathrm{the}\mathrm{position}x=0.2\mathrm{km}\mathrm{is}$$$$[\mathrm{Answer}:18{\mathrm{ms}}^{-1}]$$

Commented by Tinkutara last updated on 24/Jun/17

Answered by ajfour last updated on 24/Jun/17

$$\mathrm{for}0⩽\mathrm{x}⩽100$$$${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$$$${\mathrm{v}}_{100}^2=44+2\times 0.8\times 100$$$$=44+160={204\mathrm{m}}^2/{\mathrm{s}}^2.$$$$\mathrm{after}\mathrm{this}\mathrm{for}$$$$100⩽\mathrm{x}⩽200\mathrm{the}\mathrm{area}\mathrm{under}$$$$\mathrm{this}\mathrm{portion}\mathrm{of}\mathrm{a}-\mathrm{x}\mathrm{graph}\mathrm{gives}$$$$\int \mathrm{adx}=\int \frac{\mathrm{vdv}}{\mathrm{dx}}\mathrm{dx}=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_{\mathrm{i}}^2}{2}$$$$\mathrm{area}=\frac{1}{2}(0.8+0.4)\left(100\right)=60$$$$\mathrm{so}\frac{{\mathrm{v}}_{\mathrm{f}}^2-204}{2}=60$$$${\mathrm{v}}_{\mathrm{f}}\mathrm{is}\mathrm{the}\mathrm{velocity}\mathrm{at}\mathrm{x}=200$$$${\mathrm{v}}_{\mathrm{f}}=\sqrt{120+204}=\sqrt{324}=18\mathrm{m}/\mathrm{s}.$$$$$$

Commented by Tinkutara last updated on 24/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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