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Rasek [7]
3 years ago
14

A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

ant at 400 N during this time. Find
(a) the average power developed by the engine and
(b) the instantaneous power output of the engine at t 5 12.0 s, just before the car stops accelerating.
Physics
2 answers:
nexus9112 [7]3 years ago
8 0

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

     F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s  = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

weeeeeb [17]3 years ago
7 0

Answer:

Part a)

Power = 23850 W

Part b)

P = 47700 W

Explanation:

Part a)

As we know that the acceleration of the object is given as rate of change in velocity

a = \frac{dv}{dt}

a = \frac{18 - 0}{12}

a = 1.5 m/s^2

now we know that

F_{net} = ma

F - F_{air} = ma

F - 400 = (1.50 \times 10^3)(1.5)

F = 2650 N

now the work done by the engine

W = F.d

here distance covered by the engine is given as

d = \frac{v_f + v_i}{2} t

d = \frac{18 + 0}{2} (12)

d = 108 m

now we have

Power = \frac{Work}{time}

Power = \frac{2650 \times 108}{12}

Power = 23850 W

Part b)

Instantaneous power is given as

P = \frac{dW}{dt}

P = F.v

P = (2650)(18)

P = 47700 W

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