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

Part a)

`Power = 23850 W`

Part b)

`P = 47700 W`

Step-by-step explanation:

Part a)

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

`a = (dv)/(dt)`

`a = (18 - 0)/(12)`

`a = 1.5 m/s^2`

now we know that

`F_(net) = ma`

`F - F_(air) = ma`

`F - 400 = (1.50 * 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 = (v_f + v_i)/(2) t`

`d = (18 + 0)/(2) (12)`

`d = 108 m`

now we have

`Power = (Work)/(time)`

`Power = (2650 * 108)/(12)`

`Power = 23850 W`

Part b)

Instantaneous power is given as

`P = (dW)/(dt)`

`P = F.v`

`P = (2650)(18)`

`P = 47700 W`