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Rectilinear Motion

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?

Sample Answer

Rectilinear Motion

(a) Sketch of the Situation

[Diagram of the situation]

Assumptions:

Mercurious runs along the x-axis, with the building at the origin (0,0).
The bomb falls vertically under gravity, with its initial position at the top of the building (0,180).
Mercurious catches the bomb at the exact moment it would have hit the ground.

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(b) Time for Bomb to Hit the Ground

To determine the time it takes for the bomb to hit the ground, we can use the following equation:

y(t) = y_0 + v_0t - 0.5gt^2

where:

y(t) is the bomb’s vertical position at time t
y_0 is the bomb’s initial vertical position (180 m)
v_0 is the bomb’s initial vertical velocity (0 m/s)
g is the acceleration due to gravity (10 m/s^2)

Setting y(t) to 0 and solving for t, we get:

t = sqrt(2y_0/g) = sqrt(2*180/10) = 6 seconds

Therefore, it would take the bomb 6 seconds to hit the ground.

(c) Parameters C and D

To determine the parameters C and D, we can use the following information:

Mercurius catches the bomb at the base of the building at the exact moment it would have hit the ground.
Mercurius’s position at that time is given by x(t) = 0.

Setting x(t) to 0 and solving for t, we get:

t^3 - 36t^2 + Ct + D = 0

Since Mercurius catches the bomb at time t = 6 seconds, we can substitute t = 6 into the equation above and solve for C and D. This gives us:

C = 216
D = -180

Therefore, the equation for Mercurius’s path is:

x(t) = t^3 - 36t^2 + 216t - 180

(d) Where Mercurious Leaves the Bomb

To determine where Mercurious leaves the bomb, we can use the following information:

Mercurious reverses direction for the second time when he reaches the maximum x-value.
The maximum x-value is given by the derivative of x(t).

Taking the derivative of x(t), we get:

x'(t) = 3t^2 - 72t + 216

Setting x'(t) to 0 and solving for t, we get:

t = 4 or t = 18

Since Mercurious is at the base of the building at time t = 6, he must reverse direction at t = 18. At this time, his x-position is:

x(18) = 18^3 - 36*18^2 + 216*18 - 180 = 383 m

Therefore, Mercurius leaves the bomb 383 m to the left of the building.

Conclusion

In order to catch the bomb before it hits the ground and deposit it a safe distance from the city, Mercurious needs to run with super speed along a path described by the equation:

x(t) = t^3 - 36t^2 + 216t - 180

He will catch the bomb at the base of the building at the exact moment it would have hit the ground, and he will leave the bomb 383 m to the left of the building.

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