A car of mass 1250 kg is moving with a veloc | vedclass

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(C) The power $P$ delivered by the engine is given by $P = F_{engine} \cdot v$,where $F_{engine}$ is the force exerted by the engine and $v$ is the ve

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$1/5 \ m/s^2$

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(C) The power $P$ delivered by the engine is given by $P = F_{engine} \cdot v$,where $F_{engine}$ is the force exerted by the engine and $v$ is the velocity.$F_{engine} = \frac{P}{v} = \frac{30 	imes 10^3 \ W}{30 \ m/s} = 1000 \ N$.The net force $F_{net}$ acting on the car is the difference between the engine force and the resistive force $F_{resistive}$.$F_{net} = F_{engine} - F_{resistive} = 1000 \ N - 750 \ N = 250 \ N$.Using Newton's second law,$F_{net} = m \cdot a$,where $m$ is the mass of the car and $a$ is the acceleration.$a = \frac{F_{net}}{m} = \frac{250 \ N}{1250 \ kg} = \frac{1}{5} \ m/s^2 = 0.2 \ m/s^2$.

$A$ car of mass $1250 \ kg$ is moving with a velocity of $30 \ m/s$. If the engine delivers $30 \ kW$ of power and the resistive force acting on the car is $750 \ N$,find the maximum acceleration of the car.

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