00:06The exercise tells us, a small package is released from rest at A and moves to
00:11along the conveyor ABCD formed by sliding wheels. The package has an acceleration
00:17uniform velocity of 4.8 meters per second squared as it descends over sections A,
00:21B and C, D, and its speed is constant between B and C. If the speed of the package at D
00:27is
00:27of 7.2 meters per second, determine A, A the distance between C and D, B, A the time
00:33required for the package to reach D. To solve the exercise we set up the equation
00:37of uniformly accelerated rectilinear motion, since we have a constant acceleration at
00:42Along the length of the protractor of 4.8 meters per second, squared, we have the equality
00:48that the acceleration of A is equal to the velocity times the derivative of the velocity, that is, velocity
00:53Regarding the distance, we then solve using a differential equation with multiples
00:57separable. Velocity times the differential of B equals acceleration times the differential
01:02of X, its integrals will be defined, with limits, integral of limits, final velocity
01:07The initial part of the integrand B times the differential B is equal to the acceleration times the integral
01:12defined limits X, and X sub 0. Solving this, we obtain, clearing 2 on the second member,
01:19the equation of uniformly accelerated motion. This is demonstrative because it can be solved
01:25Without integration, without finding the previously obtained equation. Well, by solving the equation,
01:30We need to find the speed between AB, from the beginning the speed is 0, the final distance
01:35It is 3, and the initial one is 0, due to the beginning of the movement, so we have, the velocity AB,
01:41minus 0, is equal to 2 times the acceleration of 4.8 meters per second squared, times the
01:46final and initial distance. Solving, we get an answer of 5.36 meters per second.
02:02To find the distance between CD, we have the equality that distance is equal to X minus
02:07X sub 0. Initially we already have a speed of 7.2 meters per second, and the speed
02:14from BC. We substitute into the equation of accelerated rectilinear motion, solving and clearing
02:21The distance. We have a distance of 2.40 meters. We have times between AB and CD, to determine
02:39the total time. We have the acceleration equation, and we use it by solving for the
02:43times, for time AB, we replace the values found and established, the speed
02:48AB, is equal to initial 0, plus the constant acceleration, times AB, we get the time
02:54of 1.11 seconds. To find the time in CD, we have the speed
03:00D is 7.2 meters per second, which is equal to the velocity AB, 5.36, plus the acceleration,
03:07For time CD, we get a time of 0.38 seconds. Now to find the time at BC, we have the
03:14equation of the velocity, and we solve for time BC, we substitute values, we have that
03:223 meters is equal to the speed BC times the time BC, solving for BC we get the time at BC from 0
03:27.55
03:28seconds. Now we find the total time in D, it will be the sum of all the times, and the time in
03:36D is 2.06 seconds. That's all. If you found the video helpful, don't forget to subscribe.
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