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4) Given velocity as a function of time: v(t) and given Xo = 4m;-6 mls (8 mls)*sin([0.5 radlsec]*t)a) Find x(t-0 sec) = [2]x(t-3 sec) = [2]b) Find v(t-0 sec) [2]v(t...

Question

4) Given velocity as a function of time: v(t) and given Xo = 4m;-6 mls (8 mls)*sin([0.5 radlsec]*t)a) Find x(t-0 sec) = [2]x(t-3 sec) = [2]b) Find v(t-0 sec) [2]v(t-3 sec) [2]c) Find a(t-0 sec) [2]a(t-3 sec) = [2]

4) Given velocity as a function of time: v(t) and given Xo = 4m; -6 mls (8 mls)*sin([0.5 radlsec]*t) a) Find x(t-0 sec) = [2] x(t-3 sec) = [2] b) Find v(t-0 sec) [2] v(t-3 sec) [2] c) Find a(t-0 sec) [2] a(t-3 sec) = [2]



Answers

A particle accelerates from rest at a constant rate for some time moving along a straight line and attains a velocity of $8 \mathrm{~m} \mathrm{~s}^{-1} .$ Immediately it decelerates at a constant rate and comes to rest. If the total time taken is $4 \mathrm{~s}$, the distance travelled is (a) $32 \mathrm{~m}$ (b) $16 \mathrm{~m}$ (c) $4 \mathrm{~m}$ (d) $2 \mathrm{~m}$

Suppose the acceleration of a particle is given by two plus 60. Now the velocity VF t is equal to the anti derivative of the exhilaration two plus 60 DT. And this gives us two T plus six times t squared over two plus a C. What is the same? S two times T plus three T squared plus C. I suppose that the initial value For the velocity that's v. of zero is equal to four. Then we can find the value of C by plugging in 040 and equating it by four. So we get four equal to two times zero plus three time zero squared plus C. Which means that C is equal to four or 4. Our V F T is equal to to t plus three T squared plus four. How to find the distance traveled Over the interval of 0-1. We have d equal to the integral from 0 to 1 of the absolute value of v f t d T lessons for any value of T and to t plus three T squared plus four. The value of V f t is always positive. Then we can say that this is just The integral from 0 to 1 of two T plus three T squared plus four. DT integrating. We have two times t squared over two plus three times T cube over three plus four times T. This evaluated from year to one, you will get t squared plus t to the third power plus 40 Evaluated from 0 to 1 gives us onto the third tower plus one squared plus four times 1 -20, everything is zero. This gives us a value equal to six, and so the distance travel is six m.

Okay, So we have that the distance, um, of a object or particle? Um, from some fixed point, um, is given by this position function as if he is equal to t squared. Plus five t plus two. Um, t is, um, time measured in seconds. Um, and, um s f t, um, is in feet. And what we wanted you is we want to find the average philosophy, Um, for four to six seconds. Okay. And so remember, average velocity is your function evaluated at six. Minus your function, Evaluated at four. Divided by six months. Four. So this is gonna be six squared, plus five times six plus to minus all of four squared. Plus five comes for plus two, and that's gonna be divided by two. And so we should get, um, 15 feet per second. And now what we want to do is to find the average velocity from four two five seconds. And so, um, that's gonna be my function. Evaluated at five. Minus my function. Evaluated it for divided by five minus four. And so this is gonna be, um, five squared, plus five times five plus two minus all of four squared, plus five times four plus two. And then, of course, this is divided by one. And so that is going to give me a 14 feet per second. Okay. And then what we want to do is to find, um, the instantaneous ah, velocity at T equal to four seconds. Okay. And we're gonna go ahead and use the formula and let h be very small. So I'm in a evaluate. My function at four point on H is gonna be 0.0 01 minus my function. Evaluated it four all divided by that 40.1 Okay, so this is gonna be 4.1 And I miss square that plus five times 4.1 plus two, um, minus all of four square plus five times four plus two. And of course, we divide that by 0.1 And when we do, we hit 13 feet per second as the instantaneous velocity at time. Four seconds

So we're told that the distance of a particle from some fixed point can be given by the equation. S A T is equal to t square plus five t plus two. We're t is measured in seconds, and we want to find the average velocity of the particle over the time intervals going from 4 to 6 and 4 to 5. And then they want us to find the instantaneous velocity off the particle when time is able to four. So one of things will need to remember for this question is that velocity is equal to the change in our position over our change in time. So that's why we can use this average rate of change equation on S a T here. So we just need to plug everything in to our equations here. So for the average rate of change for the velocity from 4 to 6, this is going to be s of six minus s up, four over six, minus four. So it's got unplugged these in. So we're gonna have s of six. And this is equal to so 36 plus 30 plus two and adding all of that up that looks like you to give us 68. So So go ahead and replace this with 68 ends and s of war is going to be 16 plus 20 plus two, which looks like it gives us 38 so we can replace that with 38 there. So now if we were to subtract everything, So 68 minus 38 is 30 and then divided by two. That would be 15. And then what were the units were this to see if they tell us? Oh, so they don't actually say the units. We really can't put anything with it. It looks like So I was gonna say 15 and then over here from 4 to 5. So we're gonna follow the same thing that we did before. So it's gonna be s of five minus s for over five minus four. So we already know that s of four is 38. So we just need to figure out what is us of five. So us of five. So it's gonna be 25 plus 25 plus two. So that's 52. So then we would do 52 minus 38 which is 14 divided by one. So just be 14. That would be the average rate of change, or B. Now they want to find the instantaneous rate of change. At times you go four seconds. Well, we could just go ahead and follow this equation. We have appear, and I like to do this in steps. So the first thing you need identifies that are a is going to before because that's where we're interested in with the institutions rate of change. So I like toe first, go ahead and do s of so before, plus h. So it's going to plug four plus h into here So before plus H squared plus five times for plus H plus two. So expend this out. We should have 16 plus eight h most H squared plus 20 lost five h plus two. Now we can go ahead and at all the subs, we'd have 16 plus 20 plus to which is going to be 38 and then we're gonna have eight h plus five h that's gonna be plus 13 h and then lastly, we just have the H squared term right now. What we want to do for our next step is due s for plus H minus best of four. So it's for about what s a four is actually already found that over here we found s a four supposed to be 38 so we could just go ahead and plug that in directly, so there's gonna be 38 plus 13 h plus H squared minus 38 knows thes 30. It's just counts out with each other and were left with 13 h plus h squared. And then for the the last step, we're going to take the limit as h approach zero of s four plus H minus s of four all over h and kind of as a pro tip. Once you get to this point here, if, for whatever reason, whatever you plug into the numeric cannot be evenly divided by this agent the Dominator, then you may need to go back in check your ouch broke. Because at least when we're working with polynomial, they should always cancel out, and so knows here we're just left with 13 plus h, which we can apply this summer directly. Now that'll be 13 plus zero, which gives us 13. So we end up with that instantaneous, falsely being 13

Well, position is a function of time. Could be regiones integral 40 times. DT So, uh, this into girl is equal to three d squared minus 60 times DT. Since we equals Treaty Square minus 60 and solving this integral, our position is a function of time Musical two D, Q Minus free T square. Let's see where sees a constant well, it Time t is equal to zero. Uh, position is zero Q minus three to CDO square place scene and therefore si is equal to four. So C is equal to four, right? Since ah c equals all right position in Time T is equal to zero, So seize equal to four and further, like music rollup um, we have position is a function of time is equal to a cube minus three t square less four feet and position when time physical four seconds is equal to forgive Minus three into four square this fuller musical to 20 peach right, So we have that's not is equal to four feet and s one is equal to 20 feet and some office not And this one is equal to 24 feet. No, let's find out acts relation we don't which is, um TV divided Fine TT eyes equal to x relation of the function of time which is equal to 60 minus six feet or second squared. So x relation when time physical two seconds His six into two mine six Matiz ical to six beat were second square.


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