A 1520 kg car in motion rolls down a frictionless hill that is inclined at \( 14.4^{\circ} \). \( f \) the hill is 28.5 m long, what is the velocity of the car as it reaches the base of the hill? \[ \begin{array}{l}a=2.44 \mathrm{~m} / \mathrm{s}^{2} \\ \qquad v_{\mathrm{f}}=[?] \mathrm{m} / \mathrm{s}\end{array} \]

To find the velocity of the car when it reaches the base of the hill, we can use the kinematic equation: \[ v_f^2 = v_i^2 + 2a d \] Since the car is rolling down from rest, the initial velocity \( v_i = 0 \) and the distance \( d = 28.5 \, \text{m} \). Plugging in the values: \[ v_f^2 = 0 + 2(2.44 \, \text{m/s}^2)(28.5 \, \text{m}) \] Calculating it step-by-step: \[ v_f^2 = 2 \cdot 2.44 \cdot 28.5 \] \[ v_f^2 = 139.64 \] \[ v_f = \sqrt{139.64} \approx 11.82 \, \text{m/s} \] So, the velocity of the car as it reaches the base of the hill is approximately \( 11.82 \, \text{m/s} \). --- Roller coasters operate on similar principles! They start at a height, converting potential energy to kinetic energy, speeding up as they plunge down inclines. It's physics in action that creates the thrill of a ride! Want to calculate the speed of another object rolling down a different slope? Remember to take into account gravitational acceleration and the angle of the incline; just alter the angle or distance to see how physics can vary outcomes!