8) A roller coaster starts at the top of the hill at \( 0 \mathrm{~m} / \mathrm{s} \) and then races down to the bottom until it gets to \( 22.7 \mathrm{~m} / \mathrm{s} \) with an acceleration of \( 2.5 \mathrm{~m} / \mathrm{s}^{2} \). How long did it take to get from the top of the hill to the bottom in seconds?

Given: - Initial velocity (\( u \)) = 0 m/s - Final velocity (\( v \)) = 22.7 m/s - Acceleration (\( a \)) = 2.5 m/s² We can use the kinematic equation to find the time taken to travel from the top of the hill to the bottom of the hill: \[ v = u + at \] Substitute the given values into the equation: \[ 22.7 = 0 + 2.5t \] Now, we can solve for time (\( t \)): \[ 2.5t = 22.7 \] \[ t = \frac{22.7}{2.5} \] Let's calculate the time taken to travel from the top of the hill to the bottom of the hill. Calculate the value by following steps: - step0: Calculate: \(\frac{22.7}{2.5}\) - step1: Convert the expressions: \(\frac{\frac{227}{10}}{\frac{5}{2}}\) - step2: Multiply by the reciprocal: \(\frac{227}{10}\times \frac{2}{5}\) - step3: Reduce the numbers: \(\frac{227}{5}\times \frac{1}{5}\) - step4: Multiply the fractions: \(\frac{227}{5\times 5}\) - step5: Multiply: \(\frac{227}{25}\) The time taken to travel from the top of the hill to the bottom of the hill is approximately 9.08 seconds.