00:05In this video we are going to solve the following exercise, a piece of electronic equipment
00:10which is surrounded by packing material is dropped in such a way that it hits the ground with
00:14a speed of 4 meters per second. After the impact, the equipment experiences
00:20an acceleration of A equal to -K times X, where K is a constant and X is the compression
00:25of the packaging material. If said material experiences a maximum compression of 20 millimeters,
00:31According to Figure P11.19, determine the maximum acceleration of the equipment. To do this, first,
00:38We know that acceleration is equal to the constant K proportional to X. And we equate the acceleration
00:44of the velocity times the derivative of the velocity, equal to negative K times X. Now we do the integrals
00:50defined in both members, taking into account that the limits are 4 meters per second,
00:54The final and initial velocity is 0, and the definite integral also undergoes compression.
00:59of the final material of 20 millimeters after impact, and also initially it is from
01:040. Expanding the definite integrals on both sides, we have, 8 is equal to minus
01:09K times 200, this equals K, has a constant value proportional to minus 0.004, or minus
01:15a 25 to bus. Replacing the values in the acceleration, we have an acceleration of minus 0.8
01:24millimeters per second. That's all. If you found the video helpful, don't forget to subscribe.
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