Answer:
Approximately [tex]6.5 \times 10^{5}\; {\rm W}[/tex].
Explanation:
At a speed of [tex]v[/tex], the kinetic energy of the ride and the riders of mass [tex]m[/tex] (combined) would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2}\end{aligned}[/tex].
If friction is negligible, [tex]\text{KE} = (1/2)\, m\, v^{2}[/tex] would be the work required to achieve this speed. That is:
[tex]\begin{aligned} \text{work} = \text{KE} = \frac{1}{2}\, m\, v^{2}\end{aligned}[/tex].
Given that this work was completed in a duration of [tex]t[/tex], the average power would be:
[tex]\begin{aligned} & (\text{average power})\\ =\; & \frac{(\text{work})}{(\text{time required})} \\ =\; & \frac{(1/2) \, m\, v^{2}}{t}\end{aligned}[/tex].
Substitute in [tex]m = 5000\; {\rm kg}[/tex], [tex]v = 27\; {\rm m\cdot s^{-1}}[/tex], and [tex]t = 2.8\; {\rm s}[/tex]:
[tex]\begin{aligned} & (\text{average power}) \\ =\; & \frac{(1/2) \, m\, v^{2}}{t} \\ =\; & \frac{(1/2) \times 5000\; {\rm kg} \times 27\; {\rm m\cdot s^{-1}}}{2.8\; {\rm s}} \\ \approx \; & 6.5 \times 10^{5}\; {\rm W}\end{aligned}[/tex].
(Note that if acceleration is constant, the power input to the ride would be proportional to [tex]t^{2}[/tex]. The average power of input to the ride would be a quarter of the peak power input. Multiplying average velocity (proportional to [tex]t[/tex]) by average force (proportional to [tex]t[/tex]) would overestimate the average power by [tex]100\%[/tex].)
