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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-




Question Number 110926 by Aina Samuel Temidayo last updated on 31/Aug/20
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) the instantaneous power  developed by the engine when the car  reaches its final speed.
$$\mathrm{A}\:\mathrm{2000kg}\:\mathrm{car}\:\mathrm{start}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{and} \\ $$$$\mathrm{accelerated}\:\mathrm{to}\:\mathrm{a}\:\mathrm{final}\:\mathrm{velocity}\:\mathrm{of} \\ $$$$\mathrm{20m}/\mathrm{s}\:\mathrm{in}\:\mathrm{16}\:\mathrm{seconds}.\:\mathrm{Assuming}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{of}\:\mathrm{500N},\:\mathrm{find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{the}\:\mathrm{average}\:\mathrm{power}\:\mathrm{developed}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{engine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{the}\:\mathrm{instantaneous}\:\mathrm{power} \\ $$$$\mathrm{developed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{engine}\:\mathrm{when}\:\mathrm{the}\:\mathrm{car} \\ $$$$\mathrm{reaches}\:\mathrm{its}\:\mathrm{final}\:\mathrm{speed}. \\ $$
Answered by mr W last updated on 01/Sep/20
a=((20)/(16))=1.25 m/s^2   s=((20×16)/2)=160 m  F=ma+R=2000×1.25+500=3000N  (1)  P=((3000×160)/(16))=30000 watt=30 kW  (2)  P=Fv=3000×20 =60000 watt=60 kW
$${a}=\frac{\mathrm{20}}{\mathrm{16}}=\mathrm{1}.\mathrm{25}\:{m}/{s}^{\mathrm{2}} \\ $$$${s}=\frac{\mathrm{20}×\mathrm{16}}{\mathrm{2}}=\mathrm{160}\:{m} \\ $$$${F}={ma}+{R}=\mathrm{2000}×\mathrm{1}.\mathrm{25}+\mathrm{500}=\mathrm{3000}{N} \\ $$$$\left(\mathrm{1}\right) \\ $$$${P}=\frac{\mathrm{3000}×\mathrm{160}}{\mathrm{16}}=\mathrm{30000}\:{watt}=\mathrm{30}\:{kW} \\ $$$$\left(\mathrm{2}\right) \\ $$$${P}={Fv}=\mathrm{3000}×\mathrm{20}\:=\mathrm{60000}\:{watt}=\mathrm{60}\:{kW} \\ $$
Commented by Aina Samuel Temidayo last updated on 31/Aug/20
What is instataneous power please?  What is the difference between  average power and it?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{instataneous}\:\mathrm{power}\:\mathrm{please}? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{between} \\ $$$$\mathrm{average}\:\mathrm{power}\:\mathrm{and}\:\mathrm{it}? \\ $$
Commented by Aina Samuel Temidayo last updated on 31/Aug/20
What′s v? Why is it 20W?
$$\mathrm{What}'\mathrm{s}\:\mathrm{v}?\:\mathrm{Why}\:\mathrm{is}\:\mathrm{it}\:\mathrm{20W}? \\ $$
Commented by mr W last updated on 01/Sep/20
not 20 W, but the unit of 3000×20 is W
$${not}\:\mathrm{20}\:{W},\:{but}\:{the}\:{unit}\:{of}\:\mathrm{3000}×\mathrm{20}\:{is}\:{W} \\ $$

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