The velocity of an object is given by the expression v(t) =...

The velocity of an object is given by the expression v(t) = 3.00 m/s + (4.00 m/s³)t², where t is in seconds. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.000 s. A) (4.00 m/s)t B) 1.00 m + (3.00 m/s)t + (1.33 m/s³)t³ C) (3.00 m/s)t + (1.33 m/s³)t³ D) 1.33 m E) (4.00 m/s)t + 1.00 m The velocity of an object as a function of time is given by v(t) = 2.00 m/s + (3.00 m/s) t - (1.0 m/s²) t². Determine the instantaneous acceleration of the object at time t = 3.00 s. A) -3.00 m/s² B) -2.00 m/s² C) 1.00 m/s² D) 0.00 m/s² E) 2.00 m/s² In the pulley system in the figure, the pulleys and the rope are of negligible mass and there is no friction in the axles of the pulleys. If the mass of each block is m, find the acceleration of the weights and the value of the tensile force of the rope. Which object will be lifted and which one will fall? A) a? = -1/5 g; a? = 2/5 g; T = 3/5 mg B) a? = 1/5 g; a? = -2/5 g; T = 2/5 mg C) a? = 1/5 g; a? = -2/5 g; T = 1/5 mg D) a? = 2/5 g; a? = -3/5 g; T = 1/5 mg E) a? = -2/5 g; a? = +3/5 g; T = 1/5 mg

Transcript

00:01 In this problem we are given three questions.

00:06 The first two of them are related to some equations of motion in one dimension.

00:12 And in the last one we have a pulley system, a double pulley system.

00:17 So let's get started with the first one.

00:21 We are given the velocity of an object as a function of time.

00:26 V of t is equal to 3 .00 plus.

00:32 Plus 4 .00 times t squared in appropriate units.

00:40 So we are measuring t in seconds and v in meters per second.

00:47 The question is what is the position of the object as a function of time.

00:53 If it is located at x equal to 1 meters at t equal to 0.

01:01 So what is the position function? it is the integral of the velocity.

01:08 So we have the integral of v plus some integration constant, which we are going to determine using the initial value.

01:21 So the integral of the velocity function is 3 .00 t plus 4 .00 t -cube over 3 times this constant x -0.

01:36 Now we know that when t is equal to 0, x is equal to 0, x is equal to 1 .00 so setting t equal to 0 in this function of x of t we get just x 0 which directly gives the x value of x0 indeed so we can write x of t equal to 1 .00 plus 3 .00 t plus okay we have 4 over 3 3 1 .33 t cube inappropriate units of course and the first term the first constant should go like meters this should go like meters per second and this factor should go like meters per second cube so that the unit of x is just meters so this is the answer and since we are given five choices we see that the correct choice should be like that okay the next question we are again given the velocity function of some object, the velocity expression is a function of time for some object.

03:03 V of t equal to 2 .0 plus 3 .00 t minus 1 .00 t squared.

03:14 The question is, what is the instantaneous acceleration of this object at t equal to 3 seconds? so first we need to find the acceleration and we remember that it is just the time derivative of the velocity function.

03:32 So if take derivative of the expression in this line, we get 3 .0 minus 2 .00.

03:43 And we see that acceleration at t equal to 3 seconds is just minus 3 meters per second squared.

03:54 Therefore, the correct choice to this question is a.

04:05 And the last problem, we have this double pulley system hanging from some ceiling.

04:14 Let me just very quickly draw it.

04:18 So we have the first, okay, we had the second object, indeed over here.

04:24 And we have this fixed pulley.

04:27 The rope goes around a second one like this, which is movable.

04:34 Fixed at the other end and the first object is hanging from this moving pulley like this so there is no friction and there's no friction anywhere and these pulleys are messless and also the rope is the police and the rope are messless and yes there's no friction the question is if the if each of the blocks have some mass and what is the acceleration of these blocks and also the tension developing in the rope and which object will be lifted and which one will fall so let's end as the situation we are going to start with the net forces acting on these objects so since we have this gravitational acceleration in downward direction we always have this gravitational forces m g acting in a downward direction on the bodies now we have some tension developing in this rope so since we have only one rope the tension will be the same throughout this rope so let's say if there is a tensile force of t pulling the second object in the upper direction it will be carried over through this pulley and the first object will also feel the same tension but it will feel it twice because this pulley is movable and this rope is just goes around it so it will just it will be twice as effective as like moving this object in the direction.

06:52 Okay with these forces we are now going to write down the equations of motion using the newton's second law.

07:00 To do that i'm going to assume the outward direction is positive.

07:06 Okay so what are the what is newton's second equation? it is m times the net acceleration.

07:13 Okay, let's consider the first object first.

07:17 M times the acceleration of this object equal to the net force acting on this object.

07:25 With the upward direction is positive, we have the net force as 2t minus m g.

07:34 If you assume the downward is positive, you shouldn't worry about that because the signs will make up for it.

07:42 So everything is consistent as long as you follow your sign convention.

07:48 Okay, now for the second one, we have m a2 equal to the net force again.

07:57 We have t minus m g and we have two equations two linear equations but we have three unknowns we don't know t we don't know a 1 or a 2 so we need one more equation to obtain this third equation we are going to use the fact that the length of this rope is constant in time and we are going to use this fact in such a way that we are going to express the heights of these objects and we will relate these positions to the actual length of the object as follows.

08:41 Okay now suppose i imagine a hypothetical line at this level so that i know that this portion of the rope will not be moving anywhere so this portion or this height will be constant all the time.

09:03 And suppose i have the length of this portion, the this portion of the rope given by some measurement, some variable u2.

09:17 And i have some other variable to denote the length of this portion and this portion of the rope.

09:31 But i'm going to consider it twice, because we are see two pieces of ropes here and we also have some again constant portions around the these pull is like there so i'm going to call them let's say see and see as in the circumference but we are going to half of them we are going to consider the half of it but doesn't really matter as long as things are constant we will not be really interested in them so let us try to write down the length of the rope using these various various we have this constant portion h plus u1 plus c plus u1 again so i am i have done this part this part this part and this part you one now we have see again this part and u2 let us take a time derivative of the length of the rope the first term drops the second term gives the speed of this object, the first object, v1.

10:51 The second term is, the third term, is constant again, and the fourth term will give another contribution to v1...

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