Answer to Question #132068 in Mechanics | Relativity for HAKIM

Question

8. A 1.50 x 103 kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains constant at 4.00 x 102 N during this time. Find a) the average power developed by the engine and b) the instantaneous power output of the engine at t=12.0 s, just before the car stops accelerating.

Answer 1

(a) The acceleration of the car is

$a = \frac{v-v_0}{t}=\frac{18.0-0}{12}=1.5\;m/s^2$

Thus, the constant forward force due to the engine is found from ΣF = F_engine – F_air = ma as

$F_{engine} = F_{air} + ma = 400 + (1.5\times 10^3)(1.50) = 2.65\times 10^3\;N$

The average velocity of the car during this interval is v_av = (v + v₀)/2 = 9.0 m/s, so the average power input from the engine during this time is

$P_{av} = F_{engine}v_{av}=(2.5\times 10^3)(9) = 2.38\times 10^4\;W(\frac{1\;hp}{746\;W}) = 32.0\;hp$

(b) At t = 12.0 s, the instantaneous velocity of the car is v = 18.0 m/s and the instantaneous power input from the engine is

$P=F_{engine}v=(2.65\times 10^3)(18.0) = 4.77\times 10^4\;W(\frac{1\;hp}{746\;W}) = 63.9\;hp$