A 1.50 multiplied by 103 kg car starts from rest and accelerates uniformly to 19.9 m/s in 11.6 s. Assume that air resistance remains constant at 400 N during this time.

Question

A 1.50 multiplied by 103 kg car starts from rest and accelerates uniformly to 19.9 m/s in 11.6 s. Assume that air resistance remains constant at 400 N during this time.
(a) Find the average power developed by the engine.
1 W

(b) Find the instantaneous power output of the engine at t = 11.6 s, just before the car stops accelerating

drwls · Answer

(a) Paverage*t = Kinetic energy increase + work done against air friction.
KE increase = 297,000 J
t = 11.6 s
distance traveled = X = (19.9/2)*11.6 = 115.4 m

The work done against friction = 400N*115.4m = 46,170 J

Pav*t = 343,170 J
Pav = 29,584 W

The assumption of constant air friction is completly unrealistic and will underestimate the final instantaneous power level.

You can come up with an answer by using the instantaneous relation
P = (Force)*(Velocity)

For the force, add the air resistance to the force required to accelerate,
F @ t=11.6s = M*a + 400 N
= 2573 + 400 = 2973 N
Instantaneous power = F*V = 59,168 W