A car with mass of 1000.0 kg accelerates from 0 m/s to 40.0 m/s in 11.0 s. Ignor

Question

A car with mass of 1000.0 kg accelerates from 0 m/s to 40.0 m/s in 11.0 s. Ignore air resistance. The engine has a 23% efficiency, which means that 23% of the thermal energy released by the burning gasoline is converted into mechanical energy.
(a) What is the average mechanical power output of the engine?
Answer for part A is 72.7272

(b) What volume of gasoline is consumed? Assume that the burning of 1.0 L of gasoline releases 44 MJ of thermal energy.

I have the answer for part A which is 72.7272 need help finding out part B.

Explanation / Answer

(a)Let the acceleration is a m/s^2 and the distance covered in 10 sec is s meter by the car, By v = u + at =>40 = 0 + a x 10 =>a = 40/10 = 4 m/s^2 By s = ut + 1/2at^2 =>s = 0 + 1/2 x 4 x 10 x 10 = 200m Thus Power(output) = W/t = (F x s)/t = (m x a x s)/t = (1000 x 4 x 200)/10 = 80000 Watt (b)Thus power(input) =[Power(output)/Efficiency] x 100% =>power(input) = [80000/22] x 100 = 363636.36 Watt Thus Energy(input) by gasoline = P(input) x time = 363636.36 x 10 = 3.64 MJ Thus the gasoline consumed = 3.46/46 = 0.07521 L = 75.21 mL

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