A car accelerates uniformly from rest to 26.5 m/s in 8.60 s along a level stretch...

A car accelerates uniformly from rest to 26.5 m/s in 8.60 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 7.76 x 10^3 N, and (b) the weight of the car is 1.83 x 10^4 N.

Expert Solution

GIVEN

initial velocity of the car 0 m/s, final velocity of the car v= 26.5 m/s, time period is t= 8.6 s

now acceleration a= v-u/ t2-t1 = 26.5 - 0 / (8.60) = 3.081 m/s2

form equations of motion

v^2 - u^2 = 2as ,

s = v^2/(2*a)= 26.5^2 / (2*3.081) = 113.96 m

power P = w/ t = F*s /t = ma*s /t = (w/g)*a*s/t

1) given weight of the car w1= m1g = 7.76*10^3 N ==> m1 = w1/g = 7.76*10^3/ 9.8 = 791.83kg

P = 791.83 *3.081*113.96 / (8.60) =3237.9 W--------------> Ans

2) car weight w = 1.83*10^4 N , mass m= w/g = 1.83*10^4/ 9.8 =1867.34 kg

P = 1867.34*3.081*113.96 / (8.60) = 76237.577 W --------> Ans

genius_generous answered 3 hours ago

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