A-2000kg-car-start-from-rest-and-accelerated-to-a-final-velocity-of-20m-s-in-16-seconds-Assuming-a-constant-air-resistance-of-500N-find-i-the-average-power-developed-by-the-engine-of-the-car-ii-

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Question Number 110926 by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{A}2000\mathrm{kg}\mathrm{car}\mathrm{start}\mathrm{from}\mathrm{rest}\mathrm{and}$$$$\mathrm{accelerated}\mathrm{to}\mathrm{a}\mathrm{final}\mathrm{velocity}\mathrm{of}$$$$20\mathrm{m}/\mathrm{s}\mathrm{in}16\mathrm{seconds}.\mathrm{Assuming}\mathrm{a}$$$$\mathrm{constant}\mathrm{air}\mathrm{resistance}\mathrm{of}500\mathrm{N},\mathrm{find}$$$$\left(\mathrm{i}\right)\mathrm{the}\mathrm{average}\mathrm{power}\mathrm{developed}\mathrm{by}$$$$\mathrm{the}\mathrm{engine}\mathrm{of}\mathrm{the}\mathrm{car}.$$$$\left(\mathrm{ii}\right)\mathrm{the}\mathrm{instantaneous}\mathrm{power}$$$$\mathrm{developed}\mathrm{by}\mathrm{the}\mathrm{engine}\mathrm{when}\mathrm{the}\mathrm{car}$$$$\mathrm{reaches}\mathrm{its}\mathrm{final}\mathrm{speed}.$$

Answered by mr W last updated on 01/Sep/20

$$a=\frac{20}{16}=1.25m/s^2$$$$s=\frac{20\times 16}{2}=160m$$$$F=ma+R=2000\times 1.25+500=3000N$$$$\left(1\right)$$$$P=\frac{3000\times 160}{16}=30000watt=30kW$$$$\left(2\right)$$$$P=Fv=3000\times 20=60000watt=60kW$$

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{What}\mathrm{is}\mathrm{instataneous}\mathrm{power}\mathrm{please}?$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{difference}\mathrm{between}$$$$\mathrm{average}\mathrm{power}\mathrm{and}\mathrm{it}?$$

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$${\mathrm{What}}^{\prime }\mathrm{s}\mathrm{v}?\mathrm{Why}\mathrm{is}\mathrm{it}20\mathrm{W}?$$

Commented by mr W last updated on 01/Sep/20

$$not20W,buttheunitof3000\times 20isW$$

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