00:03This video presents a series of simple exercises on dynamics, the purpose of which is to
00:08To practice what has already been learned, we have the following exercise: acceleration
00:13of a particle is directly proportional to time t when time is equal to 0 the
00:19The particle's speed is equal to 16 inches per second if it is known that the speed is equal to
00:24Given that the velocity is 15 inches per second and x equals 20 inches when the time is one second, determine
00:31the speed, position, and total distance traveled when the time is equal to 7 seconds
00:36Here we have that acceleration is directly proportional to time, that is, the constant
00:41that by the time we equate to the derivative with respect to time is equal here by the time we develop
00:47the definite integrals the definite integral of limits speed and 16 is equal to the constant
00:52k by the definite integral of the upper limit time t and zero lower limit by the integrand
00:57of time t times the differential this gives us v minus 16 is equal to k times t raised to the
01:03square over 2 of
01:04limits t and zero this gives us the equation for velocity v equal to k times t raised to the power of
01:08square
01:08over 2 plus 16 and we get a speed of 15 inches per second with a time of one second
01:15Substituting, we get that k is equal to negative 2 inches per second. We also have the equation
01:21v equals k over 2 plus 16, we replace the value and obtain a speed of minus 33 meters over
01:30Second, we have the equation velocity equals 16 minus t squared. We perform this equation.
01:36The definite integral, as the problem states, is x equal to 20 inches when t equals 1 second. These are
01:44the
01:44limits of the definite integral the upper limit will be x and the lower limit will be 20 for the position or displacement
01:51Reiterating as the problem states, x equals 20 inches when t equals 1 second, so the other
01:58side of the equation when the definite integral we have t is the time and the lower limit will be
02:031 1 second solving the integrals we get x equal to 1 third times t cubed plus 16 t
02:11plus 13
02:12thirds when the time equals 7 seconds, replacing we get the position of 21 inches
02:18Also, replacing the value of 7 seconds in the speed gives us minus 33 inches per second
02:24Also, replacing the time of 0, we have an initial displacement of x sub 0 equal to 13 over 3
02:31There is also a displacement 4 x sub 4 equals one third times t cubed plus 16
02:38t plus 13 thirds
02:40We replace the time of 4 seconds, we have 47 inches, we subtract the displacements found and
02:47We obtain the final position of 87 points 7 inches. The following exercise tells us the movement of a
02:55particle is defined by the relation of x equal to 1.5 te raised to the fourth power minus 30 te
03:01raised to the
03:01square plus 5t plus 10 where x and t are expressed in meters and seconds respectively determine the
03:09position, velocity and acceleration of the particle when the time is 4 seconds for
03:15To determine the position, we start from the function x equals 1.5 te raised to the fourth power minus 30 te
03:21high
03:22squared plus 5t plus 10 where x and t are expressed in meters and seconds here we replace the
03:29value of time
03:30From 4 seconds we get minus 66 meters. To find the velocity, we perform the derivative with respect to the
03:37time, this is the derivative of t is equal to 1.5 times the derivative of t raised to the
03:42fourth minus 30 by the
03:44derivative of t squared plus 5 times the derivative of t plus the derivative of the constant of
03:5010 and
03:51We obtain the velocity function with respect to time: velocity is equal to 6 times t cubed
03:56minus 60t plus 5 where we have a time of 4 seconds, we replace and obtain a speed of
04:03149 meters per second. To find the acceleration, we take the second derivative with respect to time.
04:10This is 6 times the second derivative of t cubed minus 60 times the second derivative of t
04:15plus the
04:16second derivative of the constant 5 and we obtain that the acceleration is equal to 18 t squared
04:24minus 60 with a time of 4 seconds, replacing we get an acceleration of 228 meters per second
04:35The following exercise tells us that the motion of a particle is defined by the relation x equals 12 t raised
04:41cubed minus 18t squared plus 2t plus 5 where x and t are expressed in meters
04:48and seconds
04:49respectively determine the position and velocity when the acceleration of the particle is equal to 0
04:58To determine the speed, we start from the function x equals 12t cubed minus 18t
05:04squared plus 2t plus 5 where x and t are expressed in meters and seconds to find the speed
05:13we perform
05:13The derivative with respect to time, that is, the derivative of t, is equal to 12 times the derivative of t
05:18cubed
05:19negative 18 times the derivative of t squared plus 2 times the derivative of t plus the derivative
05:26of the
05:26constant of 5 and we obtain the velocity function with respect to time velocity is equal to 36 times t
05:33high
05:33cubed minus 36t plus 2 now differentiating with respect to time, this is the second derivative is equal to
05:4136 per
05:42the second derivative of t cubed minus 36 times the second derivative of t plus the second derivative
05:48of
05:48We set the constant 2 equal to 0, which is the acceleration, and solving for time, we get a time of 0.5
05:55Seconds, we substitute time into the equations for position, velocity, and acceleration, and we get the position of
06:033 meters, a speed of minus 7 meters per second, and its acceleration equal to 0
06:10The following exercise tells us that the motion of a particle is defined by the relation x equals 5 thirds
06:16cubed
06:17minus 5 halves of squared minus 30 themes 8 x where x is expressed in feet and
06:25seconds
06:27respectively determine the time, position, and acceleration when the velocity is equal to 0
06:32To find the speed, we take the derivative with respect to time; that is, the derivative of t is equal to 5
06:38thirds by the
06:39derivative of t cubed minus 5 halves times the derivative of t squared minus 30 times
06:45the derivative of t plus the derivative of the constant of 8
06:50and we obtain the velocity function with respect to time: velocity is equal to 5 times t squared
06:56minus 5 t minus 30
06:59where we have a speed of 0, we set it equal to 0, we factor, it's important here to obtain a positive time of 3 seconds, we get
07:07We obtain by replacing the position of minus 59.5 feet and replacing the time of 3 seconds, we obtain a speed of
07:140
07:15and an acceleration of 25 feet per second squared
07:20That's all. If you found the video helpful, don't forget to subscribe.
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