An advertisement claims that a certain \(1200-\mathrm{kg}\) car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in \(8.0 \mathrm{~s}\). What average power must the motor supply in order to cause this acceleration? Ignore losses due to friction.

To find the acceleration, we will use the formula: Acceleration (\(a\)) = (Final velocity (\(v_f\)) - Initial velocity (\(v_i\))) / Time (\(t\)) Given: Final velocity, \(v_f = 25 \mathrm{~m} / \mathrm{s}\) Initial velocity, \(v_i = 0 \mathrm{~m} / \mathrm{s}\) (since the car starts from rest) Time, \(t = 8.0 \mathrm{~s}\) Now, let's calculate the acceleration: \(a = (25 \mathrm{~m/s} - 0 \mathrm{~m/s}) / 8.0 \mathrm{~s}\) \(a = 25 / 8\) \(a = 3.125 \mathrm{~m} / \mathrm{s^2}\)

Now that we have the acceleration, we can calculate the force using Newton's second law of motion: Force (\(F\)) = Mass (\(m\)) × Acceleration (\(a\)) Given: Mass, \(m = 1200 \mathrm{~kg}\) Acceleration, \(a = 3.125 \mathrm{~m} / \mathrm{s^2}\) Now, let's calculate the force: \(F = 1200 \mathrm{~kg} \times 3.125 \mathrm{~m} / \mathrm{s^2}\) \(F = 3750 \mathrm{~N}\)

Finally, we can calculate the average power supplied by the motor using the formula: Average power (\(P_{avg}\)) = Force (\(F\)) × Distance (\(d\)) / Time (\(t\)) We can find the distance using the formula for the average velocity: Average velocity (\(v_{avg}\)) = (Initial velocity (\(v_i\)) + Final velocity (\(v_f\))) / 2 \(v_{avg} = (0 \mathrm{~m/s} + 25 \mathrm{~m/s}) / 2\) \(v_{avg} = 12.5 \mathrm{~m/s}\) Now, we can calculate the distance using the average velocity and time: Distance (\(d\)) = Average velocity (\(v_{avg}\)) × Time (\(t\)) \(d = 12.5 \mathrm{~m/s} \times 8.0 \mathrm{~s}\) \(d = 100 \mathrm{~m}\) Now, let's calculate the average power: \(P_{avg} = 3750 \mathrm{~N} \times 100 \mathrm{~m} / 8.0 \mathrm{~s}\) \(P_{avg} = 375000 / 8\) \(P_{avg} = 46875 \mathrm{~W}\) The average power the motor must supply in order to cause this acceleration is \(46875 \mathrm{~W}\).