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'3ue10Q0YA esnon walks for 10 ,0 metcrs the positive direction; then wralks for 1.51 meters tht negnnve direction; What /s the displacement0"tie Fatson me...

Question

'3ue10Q0YA esnon walks for 10 ,0 metcrs the positive direction; then wralks for 1.51 meters tht negnnve direction; What /s the displacement0"tie Fatson meters?Quesdon $espusiton of J particle moving along the * axis i5 given by * = (326t9| vhere [what is the instantaneous velocity (In mslatee

'3ue10 Q0Y A esnon walks for 10 ,0 metcrs the positive direction; then wralks for 1.51 meters tht negnnve direction; What /s the displacement0"tie Fatson meters? Quesdon $ espusiton of J particle moving along the * axis i5 given by * = (32 6t9| vhere [ what is the instantaneous velocity (In mslatee



Answers

\bullet A particle with mass $1.81 \times 10^{-3} \mathrm{kg}$ and charge of $+1.22 \times 10^{-8} \mathrm{Chas},$ at a given instant, a velocity of $3.00 \times 10^{4} \mathrm{m} / \mathrm{s}$ along the $+y$ -axis, as shown in Figure $20.58 .$ What are the magnitude and direction of the particle's acceleration produced by a magnetic field of magnitude 1.25 $\mathrm{T}$ in the $x y-$ plane, directed at an angle of $45.0^{\circ}$ counterclockwise from the $+x$ -axis?

Hi in the given problem, there is a charged particle at any instant whose position is shown here in the figure in this trigger here. This is the X axis, the via taxes and the positive set access. The charge particle is isn't here at the origin at that instant this is a positive charge particle. There is a magnetic field the bar Making an angle 45° with the positive X. Axis in counterclockwise direction and a velocity of the charged particle. At that instant we bar no other parameters given is Moss. Of the particle of the charged particle. S. M. Is equal to 1.81 into 10 days apart -3 kg. And the charge carried by this particle is Q. Is equal to Plus 1.2 Into 10 days par -8 colon. The velocity with which it is moving along via taxes Is given as 3.00 Into 10 years, par four m/s. End. The magnetic field is having a magnitude of one 25 s. La. Now we have to find the magnitude and direction of the acceleration produced in the motion of this charge particle for which we used the expression for magnetic Lawrence force experienced by the charged particle moving in a magnetic field, which is given by a bar in director of home. This is have body is equal to Q into We cross B. No. Using the right and screw rule for we cross me. The direction of this, we cross B is a long negative that axis here. This is negative that access and in the same direction there will be the acceleration. Now using the rule of the cross product visible to Q. V. B Into sign off angle between V&P. Which has given us 45°.. Or we can say this is Q. B. B into one of one. Due to what we can say. This is Q. V. By route to No. Using Newton's 2nd Law of motion. Also according to which the forces given as a product of mass of the particles vehicles acceleration. So we can say yes, I am into A. Is equal to Q. V. P by two. Or this A. Is equal to Q V. B. Bye mm. The mass of the charged particle into route to So now plugging in all known values A. Is equal to 1.2 into 10 days par minus eight kg colon. For the charge multiplied by three into Then reached the part four for the velocity multiplied by 1.25. Tesla for the magnetic field divided by mass retreat. To be 1.79 into 10 days. The bottom -1. Meet up our 2nd square, which is the answer for the magnitude of the acceleration, and it is directed along negatives, that axis in which there was a magnetic Lawrence force acting on the charge particles. Thank you.

So for this exercise were given a particles initial velocity and its acceleration and were asked a couple of things. Eso the initial velocity and acceleration are here on the screen. Um, and the first thing we're asked is at what time does the particle across the the Y axis? So let's draw a possible particle trajectory here in red. OK, let's say this is the particle particles trajectory. This is just a supposition is just in order in order to better illustrate the problem. Uh, notice that the particle crosses the y axis when the value of the X X is zero. So what we're gonna have to do is to calculate the particles trajectory and see when X equals zero. And for that point, the in that point, the particle will be crossing the Y axis. So that's right. Ah, the particles trajectory as ah usual r equals are zero plus V zero t plus de acceleration T squared over two. Um, you have a vector. And from this general formal okin are separate are into coordinates. So our is the position of the particle. We're gonna separate it in and X and y coordinates as first right the X cordant X equals x 00 is the initial position of the particle plus V zero x the the X component of the initial velocity Times T plus 80 a x the X component of exploration two squared over to okay, but just separate things here. Okay, so well, we know that the particle, the the exercise tells us that the particle starts its trajectory at the origin, so x zero equals zero. Ah, and we have the values of VI zero x. It's 11 and a X. It's minus 1.2. And we want to solve this equation for X equals zero, so we can simply do it. Uh, first, let's right here that ah x equals 11 T minus 0.60 squared, OK? And we're gonna make it equal to zero. Ah, and I'm gonna solve this equation. So let's solve it. 40 different from zero. And the solution is T Eagles 11 over 110.6, which is equal to 18.3 seconds. So the particle crosses the Y axis at this time 18 point point three seconds. Okay, this is the the answer through to the first question. Now, the second question asks us, um, what is the value of why when the particle crosses the y axis? So we're gonna have to ah, ride the same equation That's before the equation for the position of the particle. But instead of writing it for component X, we're going to ride it for a component. Why, you're gonna have, uh, why equals y zero plus V zero y times t plus A. Why time c squared over two again. Does the particle starts at the origin? So why's here? Is gonna zero v zero Why? Ah, we had that. We already have this This value it's 14 to and t the time when the particle crosses the crosses, the Y axis is 18.3. So we're gonna much multiplied by 18.3 A. Y is point 26 t square, 18 points. Three squared over to so making this calculation, we confined at the inch. The answer is approximately 300 meters. Okay, so this is the answer to oppression. Be, um and finally were asked, What is the magnitude of the velocity, the speed that the particle is moving when it crosses the y axis and at what direction? So Ah, we know that the velocity V at any time is given by the initial velocity plus the acceleration factor times the AT T, which is the time that passed since the initial velocity. Um, and we already have all these information that we have the zero Cates. Ah, right here we have a it's right here and have tea. Tea is 18.3. It's right here. It's the answer to question eight. So it was simply gonna have to were simply gonna have to plug it all in equation this equation for V and we can very easily do it. So let's do it. Uh, the first components is gonna be 11 miners, 1.2 times 18.3. Okay, I'm simply summing these two vectors, um, component by component. And then the second components, we're gonna have 14 pless 0.26 times 18.3, and the the Y direction eso the results simply the calculation is Afghanistan 10.96 I plus 18 76 j Ah. And what we're gonna have to do now is to calculate the the magnitude of the speed. And in order to do that, we're gonna take the square root off the sum of the square of the components of E. So it's gonna be 10.96 squared plus 18.76 squared. And this grew out of this is 21.73 meters per second. Okay, so this is a speed ah. Of the particle when it crosses the y axis. Uh, finally, why we have to do It's a calculate. Well, what is the direction off this? Ah, velocity vector. So again, I'm gonna draw your, uh, the velocity vector noticed that, um, according to this equation here that we found the formula for the velocity vector. Um, the Vieques components of the velocity is negative, meaning that it's gonna point to the left. And the white component of the velocity is positive, meaning that it's gonna point upwards. And then we can draw the velocity vector as this. Okay, it's pointing. It's why components is upward. It's x component is left work, and we can calculate this angle here shown in the picture in green. Okay. And this angle, the tensions of this angle is gonna be 18 points said 76 over. Ah, 10.69. So you can calculate this angle by taking the inverse tangent. The arc tangent off 18 point sits at six. Over 10.96. And doing this calculation, we get that the angle is approximately 60 degrees. Uh, no, of course. Usually we don't give the angle this angle Here. Show shown in the picture we're usually looking for this fi angle are relative to the Y x y x is Ah counterclockwise. I'm sorry, Teoh. The X X is counterclockwise here. The fight angle that showing the picture. Um, so since clearly from a picture, you can see that Phi plus data equals 180 degrees and data is 60 degrees. You can find that the angle phi equals 100 20 degrees. So the direction of the velocity vector is Ah, it's pointing at 120 degrees counterclockwise from the x X is and that is the answer to crushes. See

In this problem it has given that any self velocity is 10 Jacob meter per second. So you victories 10 Jacob and acceleration of the particle is eight IPAP plus to Jacob. So eight I kept plus to Jacob. So what will be the value of velocity vector at any instant time? T soapy vector will be caused to you victor plus a vector into t. So you vector here is 10 Jacob And the victories eight icap Plus to Jacob into T. So this is equals two 80 I cap plus 10 Plus two T. Jacob. So this is the expression for velocity vector at any time. T Okay so let us say this is a question one. Now can you find the position vector using this equation? So velocity vector can be written as the rate of change of police in vector. So D R by DT is equals two. Eight T icap list 10 Plus two T. Jacob. Therefore dear victor will be close to 80. So eight DDT. I kept less 10 plus two T. D. T. Jacob. Now we can integrate both sides. So integration of director will be close to integration of a T D. T. I cap plus integration of 10 plus two T DT Jacob. Okay So here uh at equals to zero. His art his are was zero. So at T close to T his position victories. Let us say our victor. So what will be the result of this integration? So our vector will be close to the integration of 80 will be 40 square so 40 square. And if you put the limit of upper limit so parliament will be T square and lower limiting all the cases is going to be zero. So It's need not to write those terms so better to write only the of Parliament's so 40 square plus integration of 10. So integration of 10 with respect to T will be 20. Okay uh Here uh the unit is in icap. These are capped coefficient and now we are going to write for the Jacob coefficient. So 20 plus integration of duty will be to T square by +222 will get cancelled. So the only remaining time will be the square. So plus t square Jacob. Okay, now what is the value of position vectors? So position vector is basically X I cab plus why Jacob? So this will be equals two. 40 square Icap, blessed. 10 T plus t squared Jacob. Okay now in the problem with it saying that the X coordinate of that point is 16. So it means that 40 sq is equal to X. That is equals to 16. So can we find the value of T. So we can compare these two. So if we divide both sides by four so T square will be equals to four. So T will be equals to two. Now if th equals to two then what is on comparing the coefficient of y cap? What we can say why is equals two 10 T Plus T Square. So why is equals two 20 plus T squared. So he had tears too. So it will be 20 plus two is square. that will be caused to 24. So why coordinate of that point is 24. So this is the answer for first part nine, part B. It is saying that what is the magnitude and direction of velocity of the particle at equals two. I'm sorry what is the in be part it is saying that what is the speed of the particular at that time? So we have found the value of T. Which was two. So we will use this value in this in this equation one. So, equation one was 80. Icap, velocity is 80 icap plus 10 plus to T. Jacob. Okay, the value of these two. So putting the value of the coast to to velocity vector will be 16 icap plus 14 Jacob. All right. So now we can find the value of magnitude. So magnitude will be equals two mother of the victim, which is equal to the square root of 16 square plus 14 is squared, which will be caused to 21 point To six m/s. Now, how to find direction. So direction will be equals two. 10 universe 14 divided by 16. So 10 in verse 14 divided by 16 gives 41.18 degree. So velocity is making an angle of 41.18 degree in the counterclockwise direction with the positive x axis. So, these are the answers for part B.

Hi, everyone. Tim John's here, and we're doing another problem dealing with displacement, velocity and acceleration functions. So this problem were first given the position function as s is equal 2.27 t cubed minus 0.65 T squared minus 2.35 T plus 4.4. So you can use a graphing calculator or a graphing website to go ahead and plot this function. Um, what you should see is that it will look something like this, and we're just doing this for the first five seconds and SSN feet per second and time is in seconds. So whenever time of zero, the position will start at 4.4 right here. And the question is not asking for wherever it crosses zero. So we can go ahead and put that to the side just for now. The next thing to Dio is go ahead and software the velocity, Which is this? The time derivative of the position function here. And this will go ahead and give us 0.8 won t squared minus 1.3 T minus 2.35 and velocities and feet per second. Time is, of course, still in seconds, and this is just a quadratic function. Whenever time is zero, we'll see that it's 0.235 here and it will cross zero and look something like this. We'll do this just for the first five seconds and finally plotting the acceleration. So acceleration is just the derivative of the velocity. So if we go ahead and do that, we'll get 1.62 t minus 1.30 and the derivative of a quadratic will just give us a straight line. So every time is zero. It will start here, and it will just look like a straight line crossing through the T axis right here. Acceleration feet per second squared. Time is in seconds. Yeah, and the second part of the problem asks us what is the positive time when the particle changes? It's direction. So this would be a thing would be at the exact point where the velocity is zero because the velocity would be negative. And then it would switch over to positive. And this would be a positive time that the particle will change its direction. So what we do need to do here is we need to set our velocity function here, set to zero, and then we need to solve for whichever positive time it is. So have 0.8 won t squared, minus 1.3 T minus 2.35 Okay, remember, the quadratic function is minus B plus or minus B squared, minus four A C square root and moving to a. So I have positive 1.3 plus or minus minus 1.3 squared minus four times 40.81 times minus 2.35 square root of that. And then we'll have to times 0.81 on the bottom. So go ahead and solve that. And then you should get that. The two times will get his 2.69 seconds and a minus 1.8 seconds. But we need the positive time. So we're just gonna look at 2.69 seconds. And that is the positive time that the particle changes. It's direction or the time that the velocity goes from negative to positive.


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