the-position-of-a-particle-as-a-function-of-time-is-given-by-x20-mathrmm-mathrms-tleft-30-mathrmm--2

Question

The position of a particle as a function of time is given by
$x=(2.0 \mathrm{m} / \mathrm{s}) t+\left(-3.0 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}$ . (a) Plot $x$ versus $t$ for time from
$t=0$ to $t=1.0 \mathrm{s}$ . (b) Find the average velocity of the particle
from $t=0.35$ s to $t=0.45 \mathrm{s}$ . (c) Find the average velocity from
$t=0.39 \mathrm{s}$ to $t=0.41 \mathrm{s}$ . (d) Do you expect the instantaneous
velocity at $t=0.40 \mathrm{s}$ to be closer to $0.54 \mathrm{m} / \mathrm{s}, 0.56 \mathrm{m} / \mathrm{s},$ or
0.58 $\mathrm{m} / \mathrm{s} ?$ Explain.

Nathan Silvano · Accepted Answer

So in this problem we have a function that describes the movement of a article which is minus two tea plus three teeth to the power of three. Okay, so first of all, we need to blot graph of the position of the article from the instant TV crew. Zero Open to instance, De Cruz one second. First, we need to discover the final position of the article. Anti Coast one. So let&#x27;s see. Weren&#x27;t ecos one The position is going to be minus two. Will Astri says he goes one. And when t co zero, the position is just zeros. That&#x27;s called the initial position. So before we lost the graph, we must know what is the behavior of 1/3 degree function. So let&#x27;s put an example in here. Let&#x27;s see. This is Ah, I&#x27;m Axis and this is ah, position access. We know that 1/3 degree function with a positive constant here positive it constantly here is going to be this kind of behavior here. This is the behavior of 1/3 degree function. So, since we are interested in the first part off their movement were interest in this segment in here. We&#x27;ve already know how to plot the graph. So let&#x27;s do this. We&#x27;ve dysfunction it specifically. We&#x27;re going to have I did time excess. We&#x27;re going to have the position access. And here Oh, this is position this this time when the time was one second, We are in the position. One. So the movement off this graph a lot of this graph, it&#x27;s going to be this one. Okay, so this is the answer for the first heightened. The second item I can be. You want to discover the average velocity off the particle between the times 0.15 open two times judo 0.25 seconds. So, first of all, we need to discover the position of the article in this segments in this instance. But this is going to be miners, too. Uh, times zero point 15 plus three times, little 30.15. So the third degree. Okay, so let&#x27;s calculated this after carefully this. We know that we are in position manners zero point 29 meters and in the second distance of time, we have acts. Equals minus two time. 0.25. Coolest tree times zero point 25 to the third power. Okay, so this position going to be quote too minus little point 45 meters. So the average velocity between this interval of time, it&#x27;s going to be doubt X divided by delta t. So this is simply miners, you know, point 45 plus you know, point 29 divided by little 0.25 miners. Your point 15. So what if average velocity is going to be 1.6 minus 1.6 meters per second In the next item Nazi Brighton Sea, we have to calculated again the average lost city. But this time in the interval off time 0.19 open to time 0.21. So that&#x27;s calculate this. The final position of zero initial position of 0 19 is going to be minus two time 0 19 clothes three times 0 19 to the power of three. So this initial position is going to be minus zero point 36 meters in the final position in 0.21 seconds, going to be minus two times your point went You want those trees? Times zero point went You want to the power of three. So the final position is going to be zero point 39 Minor studio. Dirty knowing meters. So finally, the average velocity it&#x27;s going to be miners Udo point 39. Plus you&#x27;re a 0.36 divided by you know 0.21 minor issue of 0.19. This is finally 1.61 minus one point six one meters per second. So respect that the average in the final item respect that the average velocity it&#x27;s closer to minors 1.62 Because the average velocity we calculated average the velocity it calculated was around 1.6 one meters per second. So it&#x27;s reasonable to suppose that the velocity and zero point

Guilherme Barros · Answer

in this problem. We&#x27;re gonna talk about velocity. Eso way need to remember is that the definition off average velocity is that V is equal to Delta acts the displacement divided by the time that the tea while the instantaneous velocity is equal to the derivative of X with respect to teach. Okay, so in our problem, we have, um a Trajectory X is a function of T that is equal to 2 m per second times T minus 3 m per second. Que take andi in question. There go is to plot this, uh, trajectory from 0 to 1. Okay, so, uh, first notice that the directory started X equals zero since x of zero is zero x of one is two minus three. That&#x27;s minus one. Also, we can find, uh, the maximum point of the trajectory. That must happen when the derivative of X, with respect to t zero on the relative is two minus three times three teeth square. So T must be who to, um, the square root off to over three. This is equal to 0.47 seconds. This is the maximum the time when the maximum happens at that time. If you substituted into our equation. We get that X is equal to 0.63 m. Okay, so and also the roots off this equation. Are they happened when two minus three t squared times? See, this is our equation is equal to zero. So the roots are t equals zero. And he is plus or minus the square root of fruit of to over three. So now I can, uh, draw a graph. We have your axes, have backs and t access and meters to use and seconds. The graph is the whole behaves as a cubic function on. It increases until it reaches 0.63 m. Something like this at this point, T is about 0.47 seconds, and then it starts decreasing. It touches the three the TXS at the square of to over three and continuous and continues to draw until, um, t equals one. This is our final point, and X equals minus 1 m. Okay, so this is our plot in question be we have to find the average velocity between the times 0.35 seconds and zero point 45 seconds. Eso the velocity will be x 0.45 seconds, minus X of zero point 35 seconds, divided by the difference of time. That&#x27;s your 0.1 seconds So V is two times 0.45 seconds, minus three times 0.45 cube minus two times 0.35 minus three times zero wound 35 cube divided by 0.1 seconds. So V is equal to 0.552 m for starters, and I have to do the same thing, but for different times. So the time the time is your 0.39 seconds in the final one is 0.41 seconds. So we is X off 0.41 minus x 0.39 divided by 0.2 So this is two times 0.41 minus three times 0.41 Cute minus two times zero point for you, Uh, 1 39 minus three times 0.39 cube divided by 0.2 So we is equal to 0.559 six meters per second and this is very close. Is your 0.6 m per second and question the we have thio say which value do we expect the instantaneous velocity at, uh T equals 0.4 seconds to be if it should be 0.54 0.56 or 0.58. Well, notice that as we are, uh, decreasing the size off our time. So basically, let me show you graphically what I mean. Notice that the first thing we did we did in question be waas to calculate the average velocity between 0.35 0.45 2nd. Okay, so the total time here is zero point one second that total time are we took into consideration. This could be taken as a new approximation off the instantaneous velocity velocity of 0.4 2nd. Because this is your 0.4 seconds is exactly at the center off this line. But notice from the definition off. Instantaneous velocity. This is DVT. Remember that the derivative is the limit of when t on absolute I&#x27;m sorry. Goes to zero off V t bliss. Absolute minus VT divided by absolute. So we want the difference of time that we&#x27;re taking into consideration to be very small. So the smaller the difference, the better. So what we did then was to consider a very small difference. Consider a difference of 0.39 0.41 now the difference was 1/5 the original business of 0.2 seconds. So we expect this to be closer to the instantaneous velocity at equals 0.4 2nd, so expect it expect instantaneous velocity to be very close to zero point 56 m percent.

Nathan Silvano · Answer

So in problem 22 we have to We have a function to describe the movement of a particle which is Let&#x27;s see, X X equals two minus five, two less tree T square. And the first item we have to go. We have to plot the graph or off the movement of this particle. When she goes from T equals zero, open to T equals two seconds. Okay, this is a lot off Ex vs see? Let&#x27;s see. So first we need to calculate what is the position with a particle 20 0 And when t goes to until you go zero particle is in the orange off the movement. So when t equals two, the position is going to be X equals minus five times two plus three times to a square square. This is it close to minus 10 plus 12. So the final position of the particle, it&#x27;s just through meters. Okay, so two meters is around here and the lot off the entire movement from T cause zero up into teak was two seconds. It&#x27;s going to be what we know. Dysfunction is the second degree function because off the G square in here and We know that in all second degree functions, the constant follows. The T square gives us the comparative depart Herbal. Since the constant, it&#x27;s positive plus tree. We have honorable christened. And this is the graph for the first item. The movement of the particle. Okay, the second item. We have to calculate the average velocity off the particle average velocity when he she goes from times Deco zero upon two times t equals one second. We know that the average velocity is defined as the difference in the position. The rider by the difference in time, the different same time is just one second the difference in position. We have to find out. Where is the final position of the particle? The same as the first? I think so. The position of the particle is going to be minus five times one plus three times one. This is a close to minus two meters. Okay, so difference in position. Find her position. Miners Danish. Your position. It is going to be minus two meters per second. This is the average velocity of the particle. And if final biting, we have to calculate the average speed of the particle. Bro this doesn&#x27;t even to calculate. We just I need to know that by definition, a function that describes the movement of a particle accelerated general function is going to be ex off t. It goes X zero. That is the initial position plus the initial velocity more to play by the time plus the acceleration divided by two multiplied by the Times Square. This is the if you go. If we compare function that we have, that is six. Sorry, It&#x27;s not a function that is minus five T plus three D square. We can see that Just comparing this through. In here we have the acceleration divided by two equals tree. So the average speed is going to be six meters per second squared. This is the final answer for this question. Thank you for watching. No. What does she want? Me. Me, Me. Me, sir. Seem correct.

Suzanne W. · Answer

the position of a particle is given by R F T. Is equal to three T squared. I had minus seven. He scored. I had -7 T cubed J hat minus five T. To the negative two K. Hat. And units of meters. What is the velocity as a function of time? So a the velocity is the derivative. So six I had -42. J. had minus 30 T. to the -3. He had in the units of meters per second. Whoa. It took my apologies, drives one more time. So the velocity is the derivatives, there&#x27;s 60. I had minus 21 T squared J. Hat plus t. 10, tease the negative three. Okay hat in units of meters per second. Then the acceleration is equal to you. six I had -42 T. J. Hat -30 teams. The negative for Okay hat and units of meters per second squared part C. What is the particle&#x27;s velocity at tables? two seconds. So we add two seconds is equal to Well, I had -84 J. Hat 1st 1.25 he had in units of meters per second. What is the speed at equals one second in two equals three seconds. So for part the you&#x27;re finding the magnitude, So the magnitude of G at 1 2nd is equal to 24 m/s and three seconds Is equal to 190 years per second. Remember it? The square root of V X squared plus Y squared plus U C squared part E. What is the average velocity between two equals 1 and equals 2? So the average velocity equals R at 2- are at one over one second, Which is equal to nine. I had -49 J. Hat Plus 3.75 he had in units of meters per second.

Nathan Silvano · Answer

So problem 23. We have a function that describes the movement of a particle such as? Let&#x27;s see X equals six T minus two T square. Okay. And the first item off the problem, we have to blot a graph for position versus time. When the particles goes from T crew zero, open to T equals two seconds. OK, so how we gonna do this First? The position when tea cozy. Here. 20 co zero article is the orange of the movements in a 2nd 1 the time equals two seconds. The position it&#x27;s going to be six times two minus two times four. So position accede Coast too. 12 minus eight. Finally, the position&#x27;s going to be for meters. Okay, so four meters. Okay, but we can put someone something around here. Time t goes to positions here. This sis, the final position of the particle. Okay, so how we gonna make the whole graph? We know that this function is a second degree function and us all second degree functions. The graph is a pardonable. And we know this because we know this is a second degree function because of the T square in here, and we know that all second will give the grief functions The constant that follows that C square gives us the concussive ity of the horrible. Since this constant is negative, the comm cover T is going to be precent. So this is the graph off this horrible for this movements. Okay, this is the first item. The second item. We want to know the average velocity and we know that average velocity is defined a difference in position, divided by different sync time. This time we have to calculate the average velocity. When the particles goes from Chico&#x27;s here upon to t equals one second. And I can we need to find out where is the position off The particle 20 equals one. So acts is going to be six times one miners, two times one. So the position is just two. Oh, sorry. Far position is for okay for meters. Fine. Let&#x27;s go. The average velocity is just a difference in position. So four miners zero divided by the difference in time. One second. So the average velocity is just for meters. First seconds and the final item. We have to calculate the average speed off the particle. The average speed before recalculate this. We just need to know that a general function that describes the movement of a particle accelerated is X of time. Is it good to X zero? That is the initial position, plus the initial velocity times time, plus the acceleration of the divided by two times D square. Okay. And just look for the function we have here we have for this problem. Exit T equals 26 two minus two T square. So we just need to compare the argument of the square, the argument of the square and the function we have. So we have here a divided by two because mine is too. So the acceleration off the average speed is just four minus four meters per second square. This is the final answer. Thank you for watching

Suzanne W. · Answer

So we have this position of a particle. We want to find the velocity at zero in one second. So for part eight The velocity is the derivative. So this is eight TI had plus six T squared. Okay. Hot Jihad drops off because it&#x27;s a constant. So the velocity and zero is equal to zero. The velocity at one is eight. I had plus six. Okay, hat. And units of meters per second? Part B. What is the average velocity between zero and 1 2nd? Yeah. O R two. Yes. All right. T one Over T 2 -1 1. That is the average velocity. Okay, So this gives us before I had Plus two K Hat in units of meters per second.

Problem

CE Predict/Explain On two occasions you accelerat…

Problem 28 Medium Difficulty

Answer

(a) See solution.
(b) $-1.632 \mathrm{m} / \mathrm{s}$
(c) $-1.64 \mathrm{m} / \mathrm{s}$
(d) Yes

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(a) See solution.(b) $-1.632 \mathrm{m} / \mathrm{s}$(c) $-1.64 \mathrm{m} / \mathrm{s}$(d) Yes

Physics

Elyse G.

Christina K.

Andy C.

Meghan M.

Math Review - Intro

Algebra - Example 1

James S. WalkerPhysics

Elyse G.

Christina K.

Andy C.

Meghan M.

(a) See solution.
(b) $-1.632 \mathrm{m} / \mathrm{s}$
(c) $-1.64 \mathrm{m} / \mathrm{s}$
(d) Yes

James S. Walker

Physics