Rectilinear Motion: The Evil Professor Mayhem Is Planning To

Rectilinear Motion The Evil Professor Mayhem Is Planning To Drop A

Indeed, this problem involves analyzing the motion of a falling bomb and the movement of a superhero, Mercurious, to intercept and manage the bomb's trajectory relative to a building. The scenario is a complex application of physics principles, involving kinematics, calculus, and algebra to determine times, positions, and parameters governing both the bomb's fall and the hero’s response.

Specifically, the task involves modeling the bomb's free fall, deriving the relevant time it takes to reach the ground under Earth's gravity, and analyzing Mercurious's pursuit along a specified path described by a cubic function of time. The problem demands finding parameters for Mercurious's position function, solving for key times such as interception and reversal points, and calculating maximum speeds, all within the context of the given constraints.

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Introduction

Understanding the rectilinear motion involved in this scenario provides insights into the application of kinematic equations, calculus, and algebra in real-world physics problems. The environment involves a projectile (bomb) in free fall and a hero moving along a pre-defined mathematical path to intercept and ultimately manage the threat posed by the bomb.

The problem is structured into six interconnected parts, each requiring detailed calculations to determine times, positions, and parameters. The main physical principles include the equations of motion for free fall, the properties of cubic functions for projectile pursuit, and maximum speed considerations.

Part (a): Sketch of the situation

The setup involves a vertical building of height 180 meters with the bomb released from its roof at time t=0. The bomb's vertical position y(t) decreases from 180 m until it hits the ground, while Mercurious starts at a horizontal distance of 864 meters from the building's base, initially aligned with the point (x,y) = (0,0), moving along a path defined by x(t) = t^3 - 36t^2 + Ct + D.

The sketch would display the building at the right, the bomb released from the top at the point (x=0, y=180), descending vertically, and Mercurious approaching from the left along a cubic path. The horizontal position x(t) indicates Mercurious's approach velocity and trajectory, constrained to x

Part (b): Time for the bomb to hit the ground

Given the initial height h = 180 meters and acceleration due to gravity g = 10 m/s², the vertical motion follows the standard free fall equation:

y(t) = y_0 - (1/2)gt^2, where y_0 = 180 meters.

Setting y(t) = 0 for when the bomb hits the ground:

0 = 180 - 5t^2 → 5t^2 = 180 → t^2 = 36 → t = 6 seconds.

Therefore, the bomb hits the ground after 6 seconds.

Part (c): Determining parameters C and D for Mercurious to intercept at impact time

Mercurious’s horizontal position: x(t) = t^3 - 36t^2 + Ct + D.

At t = 6 s, Mercurious must be at the base of the building, which we consider at x=0 (since all action occurs to the left). Thus:

x(6) = 0 = (6)^3 - 36(6)^2 + C(6) + D.

Calculations:

• 6^3 = 216,
• 36 36 = 1296, so 36 6^2 = 36 36 = 1296, making -36 36 = -1296.

So,

0 = 216 - 1296 + 6C + D → 0 = -1080 + 6C + D.

Rearranged:

D = 1080 - 6C.

For the second condition, Mercurious's position at the impact time is known (x=0), and his initial position at t=0 is at x = -864 m (since he starts 864 m away from the building from the left). The function x(t) at t=0 yields:

x(0) = 0 - 0 + 0 + D = D.

Since initial position is -864 m, D = -864.

From earlier D = 1080 - 6C, thus:

-864 = 1080 - 6C → -6C = -864 - 1080 → -6C = -1944 → C = 324.

Correspondingly, D = -864.

Therefore, the parameters are:

• C = 324,
• D = -864.

Part (d): Position where Mercurious leaves the bomb

Mercurious reverses direction for the second time when he departs from his position at which he catches the bomb (at t=6 s, x=0). To find the position where he leaves the bomb after the first reversal, we need to examine the velocity:

v_x(t) = derivative of x(t):

v_x(t) = 3t^2 - 72t + C = 3t^2 - 72t + 324.

At t=6,

v_x(6) = 336 - 726 + 324 = 108 - 432 + 324 = 0.

This indicates a reversal of direction occurs at t=6 thus confirming the second reversal occurs at this moment. The position at this moment is x(6) = 0 (as established). The second reversal occurs at the same point where he departs from the bomb's path; thus, the position where Mercurious leaves the bomb is at x=0.

Part (e): Position when the bomb finally explodes

The bomb explodes at t=24 seconds after its release; however, given the bomb's fall time is only 6 seconds, this occurs long after the bomb's impact, meaning Mercurious’s position at explosion time is computed by plugging t=24 into the position function:

x(24) = 24^3 - 3624^2 + 32424 - 864.

Calculations:

• 24^3 = 13,824,
• 24^2 = 576, so 36 * 576 = 20,736,
• 324 * 24 = 7,776,

Thus:

x(24) = 13,824 - 20,736 + 7,776 - 864 = (13,824 - 20,736) + (7,776 - 864) = -6,912 + 6,912 = 0.

Mercurious’s position at t=24 s is x=0, meaning he is back at the base of the building at the moment the bomb explodes.

Part (f): Maximum speed of Mercurious over 24 seconds and location

The speed is the magnitude of the velocity function:

v_x(t) = 3t^2 - 72t + 324.

To find the maximum, we differentiate v_x(t) to find critical points:

v_x'(t) = 6t - 72 = 0 → t = 12 seconds.

This indicates maximum speed occurs at t=12 seconds. Calculating v_x(12):

v_x(12) = 3144 - 7212 + 324 = 432 - 864 + 324 = (432 + 324) - 864 = 756 - 864 = -108 m/s.

The negative sign indicates the speed in the direction opposite to the positive x-axis (which is consistent with the reversal). The magnitude of the maximum speed is 108 m/s, and he is at position:

x(12) = 12^3 - 3612^2 + 32412 - 864.

Calculations:

• 12^3 = 1,728,
• 12^2 = 144, 36*144 = 5,184,
• 324*12 = 3,888,

So,

x(12) = 1,728 - 5,184 + 3,888 - 864 = (1,728 - 5,184) + (3,888 - 864) = -3,456 + 3,024 = -432 meters.

In conclusion, Mercurious reaches maximum speed of 108 m/s at t=12 s, positioned approximately 432 meters to the left of the building.

Conclusion

This analysis demonstrates the application of kinematic equations, calculus, and polynomial functions within the context of a hypothetical rescue scenario involving free fall and pursuit dynamics. The precise calculation of the bomb's impact time, the parameters governing Mercurious's pursuit path, and the timing of reversals offer insight into motion optimization in physics scenarios. The maximum speed calculation emphasizes the importance of velocity analysis in pursuit problems.

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