Rectilinear motion - The evil Professor Mayhem is planning

Question # 00847851 Posted By: wildcraft Updated on: 11/23/2023 04:02 AM Due on: 11/23/2023
Subject Physics Topic General Physics Tutorials:
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 1. Rectilinear motion

The evil Professor Mayhem is planning to drop a time-bomb from the top of a 180 m tall building. If the bomb hits the ground it will explode and destroy all of the new Adelaide University City. Even if it doesn't hit the ground, the bomb is set to explode 24 s after its release. The superhero Mercurious is 864 m from the base of the building, at (x, y) = (0, 0), when he sees Professor Mayhem release the bomb. In an instant Mercurious works out that, assuming that t = 0 is when the bomb is released, he needs to run with super speed along a path described by the mathematical equation, x(t) = t ^3 − 36t^ 2 + Ct + D, in order to catch the bomb before it hits the ground, turn around and deposit it a safe distance from the city, and then turn around again and return before the bomb explodes.

(a) Draw an appropriate sketch of the situation with Mercurious's position (along the horizonal) at any time t identied by the function x(t), assuming all the action takes place to the left of the building (i.e., x < 0), and the bomb's vertical position at any time t during its fall is described by the function y(t) ≥ 0.

(b) Determine the time it would take for the bomb to hit the ground if it falls under gravity with an acceleration of 10 ms^−2 .

(c) Determine the parameters C and D if Mercurious runs and catches the bomb at the base of the building at the exact moment it would have hit the ground (at which point he also reverses direction for the rst time).

(d) Determine where Mercurious leaves the bomb (at which point he simultaneously reverses direction a second time).

(e) Determine Mercurious's position when the bomb nally explodes.

(f) Determine Mercurious's maximum speed over the 24 s period. At what position(s) is he when this occurs? 

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