Question

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 constant 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.

Answer 1

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.

Answer 2

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

$P = \frac{dW}{dt}$

$P = F.v$

$P = (2650)(18)$

$P = 47700 W$