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Constant acceleration occurs when the rate of change of velocity remains the same over time. This simplifies many calculations, as you can assume a linear relationship between velocity, acceleration, and time. In the given problem, the car moves from rest to a final velocity of 13.89 m/s over a distance of 100 meters. Knowing the car starts from rest and moves with constant acceleration allows us to use the kinematic equation:\[ v_f^2 = v_i^2 + 2a s \]This equation helps determine the constant acceleration of the car, by plugging in the final velocity \( v_f = 13.89 \text{ m/s} \), initial velocity \( v_i = 0 \text{ m/s} \), and distance \( s = 100 \text{ m} \). Solving for acceleration \( a \), we find that the acceleration is approximately 0.964 m/s². With constant acceleration, we only need an initial velocity and distance to calculate the final speed.

The work-energy principle connects the concepts of work and kinetic energy, highlighting how work done on an object results in a change in kinetic energy. In this exercise, the car is initially at rest, meaning it has no kinetic energy. As the car accelerates, the engine does work on it, increasing its kinetic energy. The kinetic energy at any point can be expressed as:\[ KE = \frac{1}{2} m v^2 \]For the car in the problem, this calculates the work done on the car: 144,561.75 Joules simply through its acceleration alone. Additionally, there's work done against gravity, considering the car moves uphill against a 10% grade, climbing a height of 10 meters. This adds further work:\[ W_2 = mgh = 1500 \times 9.81 \times 10 = 147,150 \text{ J} \]Thus, the principle of work-energy helps us account for all forces acting and calculate the total work done on the car.

Power is the rate at which work is performed over time. Understanding how power relates to work and time is essential in dynamics. The formula for power is given by:\[ P = \frac{W}{t} \]In this problem, once the total work has been calculated (291,711.75 Joules), we find the time required for the car to reach its final speed utilizing the relationship between velocity, acceleration, and time:\[ t = \frac{v_f}{a} = \frac{13.89}{0.964} \approx 14.41 \text{ seconds} \]After establishing the time, the power can be easily calculated by dividing the total work by the time, resulting in the required power to be approximately 20.24 kW. This tells us the engine needs to produce at least this amount to maintain the car's motion under these specific conditions.

Kinematic equations are fundamental in solving problems related to motion, particularly in one-dimension. They relate the five kinematic variables: displacement \( s \), initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), and time \( t \). For our problem, since the motion has constant acceleration, these can be extremely useful to link known values and unknowns, such as acceleration and time.The relevant kinematic equation used is:\[ v_f^2 = v_i^2 + 2 a s \]Where the car's velocity from rest changes at a steady rate until it reaches 13.89 m/s. Moreover, to find the time taken for the acceleration to reach this velocity, the equation:\[ v_f = at \]Was employed. Mastery of these equations is crucial as they simplify calculations and help predict future positions, velocities, and times during constant acceleration.