(Gpts:) particle moves along the X-axis with acceleration given by a(t) At + 7, where mensured seconds and (position) measured[ meters If the initinl position is gi...

A particle moves along the $x$ -axis at a velocity of $v(t)=1 / \sqrt{t}$ , $t>0 .$ At time $t=1,$ its position is $x=4 .$ Find the acceleration and position functions for the particle.

So in on velocity. Could you ask Graham so we should get equal with you? Ah, listen, computing that as one less for the power off half. So what you gonna have here? And then we have one does for the off minus half on by General, we need attempts to do so. We should get thio times. We're dividing by wonders for the square root and evaluated and take up to six. So when they could just six weeks, you get nickel Too many fathers wanted if I should meet you off five not from university and it takes six. And the Aspiration I, which included the asked number prime when he called you. We we do from here. So we're gonna you'll hear Tam's wouldn't minus 1/2. And then one glass fun day a monastery on with you and village And we need to terms with the far and any anguish again it could you mind lost? Ah, far one glass for the bow off minus three on Would you? We violated under the equal to six. So we should get Geico. Germany's far 25 barrel ministry with ju. And because you're minus five far, Jennifer handkerchief I square and then we could do on the Jew. Five. Our trees are echoed. You minus one over 25. Now we have him in the form of five. Come back. So listen, we fall isn't far. And then we should have a go to when it's far on the 1 25

3.6 Problem number 87. So 3.6 Problem number 87. So, in this case, they give us the formula for the position of a particle. So s is equal to the square root of one plus for T. Yes, and they ask us to find the velocity and acceleration at eagle one. So I need to find the velocity and the acceleration at T equal to one. So to find the velocity that's going to be the first derivative. So D s d t is going to be won over two square root of one plus for tea times four. So that is just simply to over one plus 40 square root. And we want to evaluate that when t is equal to one. So when t is equal to one, um, this is one. Excuse me. Valya, wait. Sorry. Let me back out. This was T quota six. I had a type of here. Evaluate this when t is equal to six. Not T. What? One. So when t is equal to six, this is going to be too over the square root of 25 which is 2/5. And if I look at my original problem. I think this was stated in meters per second, so the velocity would be in meters for a second. What now if I look it to find the acceleration To find the acceleration, I'm gonna have to find the second derivative of s with respect to t. Okay. And so that is going to be two times negative to one plus four t to the three halves times four. And so this becomes negative for over one plus four. T to the three has power, and we want to evaluate that when t is equal to six. So when tears equip the six, this becomes negative four over. So this becomes, what, 25 to the three halves. So that is five cube. So this is negative four over 1 25 and that's meters per second squared. So the velocity at equal six is 2.5. Should be to Fitz and the acceleration negative for 1 25th

Section three That six problem over 101 So they give us a position function that shows the position of a particle moving along a coordinate line is s equals squared of one plus 40 where we've got t measured in seconds and s measured in meters rest to find the velocity and the acceleration. So the velocity by definition, is just the change of position over time. So ds d t so that is going to be 1/2 one plus 40 times The derivative of one plus 40 is just four. So this is just too divided by the square root of one plus 40 were asked to evaluate this. When t is equal to six, that gives us two over the square root of 25. So that is just 2/5. And the units for that are in meters per second. So the velocity of this particle is 2/5 meters per second. Now we're gonna ask to find the acceleration, so the acceleration is going to be the second derivative DST t squared. So we already found that d S d t Woz what to be have to over Route one plus 40 or we might want to write this as to time one plus 40 to the minus 1/2. So how do we find the second derivative of that? Where that's going to be minus 1/2 times, two times one plus four t to the minus three halves times the derivative of one plus 40 which is just four. And so this becomes minus four over one plus for tea to the three have power. Then we just need to evaluate this. AT T equals six. We evaluate that AT T equals six. You get minus four over. So you six. That's when the 25 so 25 to the three halves. So that's five cubes. So this is negative four over 1 25 and this is measured in meters. Uh, her second squared is the acceleration. So we found the velocity and the first step, and then we differentiated again to find the acceleration, and we evaluate each at the value of T equal to six

So in this problem we have the following conditions. So initially a particle is moving an initial velocity a piece of all and started at the position Sfo from the origin and moving at a constant acceleration. A now they know that if a particle is moving at a constant acceleration then its distance with respect the time is being obeyed by this equation right here, SFP is equal to one half a t squared. Just piece of OT does SFO we're in, as I mentioned as oppose the initial position, invisible is the initial velocity and they is the constant acceleration. If the, if the acceleration is not constant then this equation cannot be used. Now the animal for this problem is to identify what the values of a peaceful and sfor if after one half second Or as of 1/2 The distance traveled is seven If after 1 2nd or s of one, the distance traveled is 11 and after TC called three halves seconds. The distance traveled is 17 uh meters. The units are not mentioned here, but just using the assumptions that this equation uses the S. I. Unit, so I'm assuming time is in second and this is is immediate. Okay, so what we're going to do here is that we substitute each of the following conditions. So at a time musical 1/2 as of T is equal to serve it. Using the equation we have seven is equal to 1/2 the If I am to finish 1/2 square that These are all times 1/2. I'm just plugging in the values of T and S. F. T. In the equation. So here will say 76-1 over eight a Plus one have a piece of old that S. M. O. And let me say SFO is equal to one of 1/8. Sorry that several is equal to seven -1 over 8.8 -1/2 is a growth. And let me call it equation number one Time is equal to one is of the is serving sorry east 11. So we have 11 here or SFP and then plugging in the time on the equation we have one half eight times one square. This piece of all times one plus is a world. So we'll end up having 11. Support to one have a less visible. The SFO. And let me call it equation number two. Now when time is equal to three hubs SRT is 17. So again these values to the equation we have 17 people to one have eight times a time of three hubs square. That First piece of old times three hubs that's possible 17 here is equal to Um 9/8. So they have squared is 9/4 times one half. So that should be 9/88 plus rehabs. Visa gold. No actually well And let me call it the question # two. Okay so what I wanted to do now is to substitute equation number one two, equations 2 and three. So that I can minimize the number of variables. So substitute one. The equation #2 Let's try to do that. We'll have 11- is equal to 1/2 a. That's visible. Plus. sfo is equal to seven -1/8 a -1/2 the Circle. And I think that we can combine all the A's and the B's of those variables. Here we have 11 is equal to one of eight minus 1. 8/8. Let's try to do that. So one half -1/8 a. We'll end up having three H. Probably. And then we have bees of all mines 1/2 piece of also will have Plus 1/2 piece of all 27. We can combine seven with the 11 on the other side so that will be 11 -7. So we'll end up having four is equal to 3.8/8. This 1/2 visa growth. Yeah let me call that equation number course. Yeah. Now can we substitute equation one two, Equation # three. So we have 17 People to 9/8. A. Those 3/2 piece of all the SFO generated from conclusion number one which is seven nine. This 1/80 -1 half the symbols combining all the A. And visible terms. We have 17 people do 9/8 a minus 1/8. 8 is 8/8. Or simply a Now 3/2 piece of old -1/2 piece of all is to over two visible or simply visible. The seventh Combining 7-17 on the other side will have 10 A sequel to eight plus peaceful. And let's call it equation number five. Um Right away how about we changed this one? This equation a little bit and let's say then minus A is actually equal to peaceable. Okay. And let's call it a question # five. So what I will do here is I will substitute equation number five. The equation # four. So that Question # four will end up having just valuable a. That being said so question number five situation number four We end up having four. You see what the 3/8 a. That's 1/2 piece of all which is equal to 10 -8. So that should give me a value of four is equal to 3/88. Just 51 halftime stain is five minus one half times eight is minus one half a combining AIDS. And the constant values. So we'll have -1 on the other side. 3/8 minus one half for the east. We'll have negative 1/8 E. Clean and multiplying both sides by negative aid to eliminate that negative 1/8 we'll end up having value of a equal to possibility. So this is not the value of T. Mhm key. Now we are we are also interested to find out what piece of oh and S. F. O. Is. So let me erase this portion of my solution to keep space for my calculation for visible and S abroad. So let's first calculate visible from equation number five. It says peaceable You see for the 10- T. So that being said visible is 10 -8 for a visible is supported too, and finally again sold for SFO from equation number one, which is equal to seven -1/8 of a, which is eight -1 half of the servo, which is to, So sfo is simply equal to 7 -1 -1, or sfo is equal to sorry, and that should be the values of these initial conditions variable for this given problem.