Answer:
Step-by-step explanation:
With a factor of (t - 1) we know that zero (ground level) is reached at 1 second from an initial height of (0 - 1)(0 - 1)(0 - 11)(0 - 13)/3 = -1•-1•-11•-13 / 3 = 47⅔ meters at t = 0
As we have <em>two </em>factors of (t - 1) we know the track does not go underground at t = 1, but rises again.
At t = 11 seconds, the car has again returned to ground level, but as we only have a single factor of (t - 11) the car plunges below ground level and returns to above ground level at t = 13 seconds due to the single factor of (t - 13)
we can estimate that the car is the deepest below ground level halfway between 11 and 13 s, so at t = 12. At that time, the depth will be about (12 - 1)(12 - 1)(12 - 11)(12 - 13) / 3 = -(11²/3) = - 40⅓ m.
we can estimate that the car is the highest above ground level halfway between 1 and 11 s, so at t = 6s. At that time, the height will be about (6 - 1)(6 - 1)(6 - 11)(6 - 13) / 3 = 5²•-5•-7 / 3 = 291⅔ m.
It's obvious that the roller coaster car had significant initial velocity at t = 0 to achieve that altitude from an initial height of 47⅔ m
