Question

a roller - coaster is at the top of a 54 - meter hill. the car and its passengers have a total mass of 1,063 kilograms. by the time the car reaches the bottom of the hill, its speed is 74 miles per hour (33 meters per second). how much kinetic energy does the car have at the bottom of the hill?
$ke = \frac{1}{2}mv^{2}$ (1 point)
2,910,494.0 j
29,403.0 j
578,803.5 j
592,416.0 j

Explanation:

Step1: Substitute values into formula

Given \(m = 1063\space kg\), \(v=33\space m/s\), and \(KE=\frac{1}{2}mv^{2}\).
Substitute \(m\) and \(v\) into the formula: \(KE=\frac{1}{2}\times1063\times33^{2}\).

Step2: Calculate \(33^{2}\)

\(33^{2}=33\times33 = 1089\).
So the formula becomes \(KE=\frac{1}{2}\times1063\times1089\).

Step3: Calculate \(\frac{1}{2}\times1063\)

\(\frac{1}{2}\times1063 = 531.5\).
Now \(KE = 531.5\times1089\).

Step4: Multiply the two numbers

\(531.5\times1089=(500 + 31.5)\times1089=500\times1089+31.5\times1089\)
\(500\times1089 = 544500\), \(31.5\times1089=(30 + 1.5)\times1089=30\times1089+1.5\times1089=32670+1633.5 = 34303.5\)
\(544500+34303.5=578803.5\space J\)

Answer:

578,803.5 J