Question

-- - - -- --- - Pmax 2nd gear 3rd 4th 5th Problem 1: The engine of a 33001b automobile provides maximum power of Pmax=200hp. During an acceleration test the power, P delivered by the engine is approximated by the adjacent histogram. a) Calculate the car's velocity (in mph) at 10 seconds if initially at rest. [5] 1 2 3 4 10 t(sec) 10 t (sec) b) If during 0<t<2s the acceleration can be assumed constant, determine the velocity of the car as a function of time, t and plot the function v(t) Vst. (Plot means using plotting software, e.g excel, matlab, etc.) [5]

Answer 1

is per The area under Porrer time graph to the vehicle [ Porrer freagy. This Energy = kinetic frergy of the will indicate how much energy By conservation of energy for vehicle [as initial K. E=o as the was at rent] vehicle The area under P-t graph from to to talonumas = [Area of 5 triangles & Area of one rectangle] * P may { [2x} + 1x H+1x1+1x1 +1x17 +(00-6)xt} = {{[1+1+1+1+1] +4} Prowy Press - 1 / 2 + 4) = = 6.5 Pimax Pman = 200 HP = 200* 0.746 Kw = 200x TAGEN V 49200 m2 3300 xyz mis ST 14 9 200 W mv 149200 [m = 3300 16] - 14.12 mis 17710 - ft-145 550 * 32*2 200 Wp = 200 x 11310 British S lupe 550 [ifflux 200 x2 1 ½ x 3300 *V2 = 1770 tls. = 46*332 7 VE = X200 3300 This - Velvaly at the end of to sunds 146 332 ftis - 461 332 × 36000 mph 1760x3 = 31.59 ron [29473 mph | the velocity is for constant given by acceleration, = urat equation mining the Na final velouty na Initial velocity az constant accordan. Energy upo 2 sunds = { x2 *1/₂ max = Prove 200 x 17 710 = 177 low -{mv. = 17.71000 = ~ = 2x13.710w = 32.762 SHS . 32.762 = a +a (2) > a = 32.762 _ 16 341 fors? (from 0 to 2 seconds) = 16 361€ [ van fils t in sunds * Taken g= 32 2.fr/52 149 29 SP unt mu - 1492000 & 2 = 14g 2 ms mis - 9-9837 mis

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