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The position of particle is given in cm by x = Find the maximum speed. mfscos 8ut, Where is in seconds.(b) Find the maximum acceleration of the particle _ m/s2(c) W...

Question

The position of particle is given in cm by x = Find the maximum speed. mfscos 8ut, Where is in seconds.(b) Find the maximum acceleration of the particle _ m/s2(c) What is the first time that the particle is at xand moving in the ~X direction?

The position of particle is given in cm by x = Find the maximum speed. mfs cos 8ut, Where is in seconds. (b) Find the maximum acceleration of the particle _ m/s2 (c) What is the first time that the particle is at x and moving in the ~X direction?



Answers

The equation of motion of a particle started at $t=0$ is given by $x=5 \sin (20 t+\pi / 3)$, where $x$ is in centimetre and $t$ in second. When does the particle (a) first come to rest (b) first have zero acceleration (c) first have maximum speed?

So we're told based on experimental observations, the acceleration of a particle is defined by the relationship. A equals minus the quantity is your 0.1 plus sign of X over B where a and X are expressing meters per second squared meters. Um were given B 0.8 meters and that the velocity is one meters per second when x zero and we want to determine the velocity off the particle when X equals minus one meters and then the position where the velocity is maximum and the maximum for us. All right, so here's our, um, expression for our acceleration. And again we could put x double dot here and try to solve differential courage in, but that would be very ugly. So we'll use our integration method. And we know that, um, uh, X equals zero philosophy is one meters per second. So you can use that for our initial state here and then we're gonna in a integrate to some unknown velocity, have the prime TV prime dummy variable here. And then we have our, um, initial the our acceleration. And again we should use a dummy variable here. Um, and we're going to go from zero to summon no value X so you can do those inner girls. And we get that we get 1/2 times V squared, minus one equals buying a 0.1 x times. Um, be times plus B times co sign of X over B minus one. Okay, um and then we can figure out put in X equals minus one. And if you got what you want is since we're given be and we get that V one is either plus or minus, um, 0.323 meters per second. So we're not actually sure which way it's going. Um, but I would assume that it's decelerating, so it's gonna be at the positive value here. Okay, so then we can figure out when V is a maximum. And that happens when a here is zero on and, um or when devi the x zero. Because this'll is not zero, assuming this is not zero. Um, so that gives us that went sign of X one over B equals minus. You appoint one. So this is the position where the velocity is maximum, and we get that That is about minus zero point 080 meters and we can plug that back in to our expression for V in terms of X and get the max is that is one point 004 meters per second. And then we can actually make a plot again of how this thing behaves. And we see that we air that, um initially were when X equals zero s o x zero. Here we have a velocity off one meter per second. So we start off here and then we come this way, and then we get our maximum displacement. Um, and we then get our all right site. Yeah, I maximum displacement. And then we come back down here and we hit a minus. Um, the velocity when, um, X equals zero again. But now coming back up. And now we come over here and we get our, um, maximum value of X in the negative direction. And we keep recycling around this thing. So we start here yet, um, equals zero. We come back, X equals zero again, and now we come back up. And so we're actually just moving back and forth. Um, as as we move around here in time

Motion in a straight line. The question is we have the expression for the position that is nine days squared minus. Thank you. Right. We have to find the position of the particle when it achieves the maximum spirit along the positive X. Direction. So we know that the speed would be equal to the expediency. Right? So when we differentiate this, we get 18 B minus ladies. But the velocity or the spirit would be maximum when David by DT Yes, he wants to judo. Right. Davey by Davey would be equal to 18 -90. So we can equate this to judo. So the 18 -60. We get the value of time as three seconds. Right? The position of the particle at three seconds would be equal. Do nine multiplied by the square minus. Thank you. That comes out to be 54 m. Right? So the position of the particle will be 54 m when they experience maximum. Hence the option, is that correct?

In this question we learn about some basic concept of the explosion. Like the person is The expression for the position is given that is 90 nineties where minus day cube night. Yeah so we have to find the position of the particle when it itches. Which is the maximum speed along the positive X. Direction. As we know that speed would be. Well do the X. Y. D. D. Right so when we differentiate this we get 18 t minus ladies. Good now to maximize developed city we have to equip David by D. Day Is equal to zero for maximum. Right now the differentiation of velocity or the differentiation of his spirit would be 18 minus 60. This has to be equated to judo. So this will give us the time as three seconds. So I B equals the three seconds the particles it just the maximum speed. Now the position of the particle At T. equals the three would be nine multiplied by a place with minus. Thank you. That is equals to 54 m. Mhm. Right so the position of the particle will be 54 m. Hence the correct option is a

Question 1 13 Tradition function is extra. Plate five. He is glad minus for thank you. The velocity is equal to that differentiation. Off the position function B X by they date. This is a dude then be minus Well, these were the escalation. Is that differentiation Off the velocity D V by they be. This is a courtroom, Ben minus 24. Be at equal do two seconds. The value of the velocity is we quit then in the dough minus well into four. This is a Quito minus. Gonna be it. Me that. But a second at equal to two seconds. The acceleration is we could do 10 minus what we ate equal to minus 38 meter purse. Again. This well bite See form Exuma off the position function. We will differentiate the position function and equate it to zero. Are then be minus. Well, the square is equal to zero. This will give us the value off B as zero point it three three said guns. At this value off time, the velocity becomes Tzeitel and the maximum value off the position is exit. Quick boom by In the 0.833 is square minus foreign toe. Zero point. It reads the you solving this, we will get the maximum value of the position as 1.1 states meet us.


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