### Damping Mechanism of a Triple-Spring Bumper System**Problem Statement:**A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass which is traveling at 45 m/s as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the compression exceeds 0.7 m as shown on the graph. Determine the maximum compression $ x $ of the outer springs.**Illustration Description:**The setup comprises a large mass approaching the bumper system at a velocity of 45 m/s. The bumper system includes an arrangement of three springs - two outer springs and a central spring. The central spring engages its deceleration effect when the compression of the system exceeds 0.7 meters.A deceleration graph is provided:- **X-axis (horizontal)**: Represents the compression, $ x $, in meters (m).- **Y-axis (vertical)**: Represents the deceleration, in meters per second squared (m/s²).**Graph Details:**The deceleration graph is linear, with the following key points:- At **0 to 0.7 meters compression**: - The deceleration is constant and remains at 0 m/s².- At **0.7 meters compression**: - The deceleration begins to increase.- At **1.4 meters compression**: - The deceleration reaches approximately 4380 m/s². The graph implies a direct proportionality between compression beyond 0.7 meters and the rate of deceleration.**Objective:**To find the maximum compression $ x $ of the outer springs, as experienced by the mass traveling at 45 m/s.**Conclusion:**The answer box is provided for students and users to input their calculated value of $ x $ (in meters), indicating the maximum compression of the outer springs as affected by the motion of the mass.---**Answer:**\[ x = \_\_\_\_ \text{ m} \]

The initial deacceleration is a=-1460-00.7-0xa=-2086 x The acceleration is equal to a=vdvdx…

Answered: A bumper, consisting of a nest of three…

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