Theme Park Physics: Calculating Velocity at 4.5 g

[schan11]

I really need help with this question.

The ‘Tower of Terror’ is a 400 m track that stretches for 300 m horizontally before curving
upward for 100 m . A 6 tonne pod with 16 people aboard (total mass about
7000 kg) is accelerated from rest (point A) to 160 km/h (at point B) along the horizontal
section by electromagnets that draw 2.2 megawatts for 6 s. After this the pod goes
unassisted into a vertical curve of radius 100 m, which gradually tightens to a curve of
radius 50 m (point C) before traveling vertically for the last part of the trip.
By this stage 12 s has elapsed (point D). In another 12 s the pod will be back to the start.




At the top of the curve (point C), the centripetal acceleration is 4.5 g. Calculate
the velocity at this point.

 

[kuruman]

A lot of numbers are provided, but only two count as far as finding the velocity at point C, the centripetal acceleration ##a_c## and the local radius of curvature ##r##.
$$a_c=\frac{v^2}{r}~\rightarrow~v=\sqrt{a_c~r}=\sqrt{4.5\times 9.8~\rm{m/s^2}\times 50~\rm{m}}=47~\rm{m/s}$$
Answer: The velocity at point (C) is 47 m/s straight up.