A 1600 kg vehicle moves with a velocity of 19.5m/s Calculate the power required to reduce ice the velocity to 3.20m/s in 11.0 s. 250 words remaining square

To calculate the power required to reduce the velocity of the vehicle, we need to follow these steps:<br /><br />1. Calculate the change in kinetic energy of the vehicle.<br />2. Calculate the work done to change the kinetic energy.<br />3. Calculate the power required to do the work in the given time.<br /><br />Step 1: Calculate the change in kinetic energy of the vehicle.<br /><br />The formula for kinetic energy is:<br /><br />\[ KE = \frac{1}{2}mv^2 \]<br /><br />where \( m \) is the mass of the vehicle and \( v \) is the velocity.<br /><br />Given:<br />- Mass of the vehicle, \( m = 1600 \, \text{kg} \)<br />- Initial velocity, \( v_i = 19.5 \, \text{m/s} \)<br />- Final velocity, \( v_f = 3.20 \, \text{m/s} \)<br /><br />Calculate the initial and final kinetic energies:<br /><br />\[ KE_i = \frac{1}{2} \times 1600 \, \text{kg} \times (19.5 \, \text{m/s})^2 \]<br />\[ KE_i = 800 \times 380.25 \]<br />\[ KE_i = 304200 \, \text{J} \]<br /><br />\[ KE_f = \frac{1}{2} \times 1600 \, \text{kg} \times (3.20 \, \text{m/s})^2 \]<br />\[ KE_f = 800 \times 10.24 \]<br />\[ KE_f = 8192 \, \text{J} \]<br /><br />Calculate the change in kinetic energy:<br /><br />\[ \Delta KE = KE_i - KE_f \]<br />\[ \Delta KE = 304200 \, \text{J} - 8192 \, \text{J} \]<br />\[ \Delta KE = 295008 \, \text{J} \]<br /><br />Step 2: Calculate the work done to change the kinetic energy.<br /><br />The work done is equal to the change in kinetic energy:<br /><br />\[ W = \Delta KE \]<br />\[ W = 295008 \, \text{J} \]<br /><br />Step 3: Calculate the power required to do the work in the given time.<br /><br />Power is defined as the work done per unit time:<br /><br />\[ P = \frac{W}{t} \]<br /><br />where \( t \) is the time.<br /><br />Given:<br />- Time, \( t = 11.0 \, \text{s} \)<br /><br />Calculate the power:<br /><br />\[ P = \frac{295008 \, \text{J}}{11.0 \, \text{s}} \]<br />\[ P = 26800 \, \text{W} \]<br /><br />Therefore, the power required to reduce the velocity of the vehicle from 19.5 m/s to 3.20 m/s in 11.0 s is 26800 W.