Question

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, which lies 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 head back before the bomb explodes. (a) Draw an appropriate sketch of the situation with Mercurious's position (along the horizontal) at any time $t$ identified 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 $y(t) \ge 0$. (b) Determine the time it would take for the bomb to hit the ground if it falls under gravity with an acceleration of $10 \text{ms}^{-2}$. (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 first time). (d) Determine where Mercurious leaves the bomb (at which point he simultaneously re- verses direction a second time). (e) Determine Mercurious's position when the bomb finally explodes. (f) Determine Mercurious's maximum speed over the 24 s period. At what position(s) is he when this occurs?

          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, which lies 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 head back before the bomb explodes.
(a) Draw an appropriate sketch of the situation with Mercurious's position (along the
horizontal) at any time $t$ identified 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 $y(t) \ge 0$.
(b) Determine the time it would take for the bomb to hit the ground if it falls under
gravity with an acceleration of $10 \text{ms}^{-2}$.
(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 first time).
(d) Determine where Mercurious leaves the bomb (at which point he simultaneously re-
verses direction a second time).
(e) Determine Mercurious's position when the bomb finally explodes.
(f) Determine Mercurious's maximum speed over the 24 s period. At what position(s)
is he when this occurs?

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, which lies 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 head back before the bomb explodes.
(a) Draw an appropriate sketch of the situation with Mercurious's position (along the
horizontal) at any time t identified 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 first time).
(d) Determine where Mercurious leaves the bomb (at which point he simultaneously re-
verses direction a second time).
(e) Determine Mercurious's position when the bomb finally explodes.
(f) Determine Mercurious's maximum speed over the 24 s period. At what position(s)
is he when this occurs?

Added by Steven L.

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