A car of mass 1250 kg experience a resistance of 750 N when it moves at 30ms⁻¹ . If the engine can develop 30kW at this speed, the maximum acceleration that the engine can produce is

The correct option is 0.2ms⁻².

Given Data:
Mass of the car, m = 1250 kg
Resistive force, R = 750 N
Velocity of the car, v = 30 ms^-1
Power developed by the engine, P = 30 kW = 30,000 W

Step 1: Calculate the driving force exerted by the engine
The relation between power, force, and velocity is given by:

$P=F\times v$

Rearranging the formula to find the driving force ($F$):

$F=\frac{P}{v}$

Substituting the given values:

$F=\frac{30000}{30}=1000\text{N}$

Step 2: Calculate the net force acting on the car
The net force ($F_{\text{net}}$) moving the car forward is the driving force minus the resistive force:

$F_{\text{net}}=F-R$

$F_{\text{net}}=1000-750=250\text{N}$

Step 3: Calculate the maximum acceleration
According to Newton's second law of motion:

$F_{\text{net}}=m\times a$

Solving for acceleration ($a$):

$a=\frac{F_{\text{net}}}{m}$

$a=\frac{250}{1250}=0.2{\text{ms}}^{-2}$

Therefore, the maximum acceleration that the engine can produce is 0.2ms⁻².