18 A car of mass 1800 kg accelerates along a horizontal road so that its speed increases from 20ms^-1 to 25ms^-1 in a time of 5.4s. What is the average useful power output of the car's engine? A 4.2 kW B 38KW C 120kW D 1100 kW

To find the average useful power output of the car's engine, we can use the formula for power:<br /><br />\[ P = \frac{W}{t} \]<br /><br />where \( P \) is power, \( W \) is work done, and \( t \) is time.<br /><br />First, we need to calculate the work done by the car. The work done is equal to the change in kinetic energy of the car:<br /><br />\[ W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]<br /><br />where \( m \) is the mass of the car, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.<br /><br />Given:<br />- Mass of the car, \( m = 1800 \, \text{kg} \)<br />- Initial velocity, \( v_i = 20 \, \text{m/s} \)<br />- Final velocity, \( v_f = 25 \, \text{m/s} \)<br />- Time, \( t = 5.4 \, \text{s} \)<br /><br />Now, calculate the change in kinetic energy:<br /><br />\[ \Delta KE = \frac{1}{2} \times 1800 \times (25^2 - 20^2) \]<br />\[ \Delta KE = \frac{1}{2} \times 1800 \times (625 - 400) \]<br />\[ \Delta KE = \frac{1}{2} \times 1800 \times 225 \]<br />\[ \Delta KE = 202500 \, \text{J} \]<br /><br />Next, calculate the average power output:<br /><br />\[ P = \frac{202500}{5.4} \]<br />\[ P \approx 37593.15 \, \text{W} \]<br /><br />Therefore, the average useful power output of the car's engine is approximately 38 kW.<br /><br />So, the correct answer is:<br /><br />B. 38 kW