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Problem 3: If an object moves with v = 3t,and =2 when =0, find the equation of motionf(t)-...

Question

Problem 3: If an object moves with v = 3t,and =2 when =0, find the equation of motionf(t)-

Problem 3: If an object moves with v = 3t,and =2 when =0, find the equation of motion f(t)-



Answers

If $s=\frac{1}{2} t^{4}-5 t^{3}+12 t^{2},$ find the velocity of the moving object when its acceleration is zero.

Hello. So if an object is traveling along the X axis such that its velocity at a time but 80 seconds is given by the is equal to T plus two three per second. Then we're trying to determine how far it's going to travel between while T. Equals zero and T equals one seconds. So we know that if an object we're moving at a constant velocity that the distance traveled is just going to be D. Is equal to V. Times delta T. Now if we think about the problem graphically the distance traveled is equivalent to the area under the plot of the velocity with respect to time since our velocity here is not constant. We first must subdivide the curve into end pieces and then let n tend off to infinity. So first let's calculate the width of each rectangular subdivision. That's going to give us delta T. Here which is going to be equal to b minus a over Ennis. That's going to be equal to one minus zero over end which is just equal to one over end. Next let's calculate index of each circumscribed rectangle. Using our curve Y that's gonna give us that T survive is equal to while I times delta T. Which is equal to just I over end. And then we calculate the distance of each piece. Using the velocity function V is equal to X plus two. So we have V. Of T. Sub I times delta T. Is going to be equal to one over N plus two times one over end. That gives us a two N plus I over and squared. Okay now we can go ahead and calculate our some. So we get our area A of our suburban is going to be equal to F of T. One times delta T plus fft two times delta T. Up to plus F of T N times delta T. That's going to give us here A two N plus one over N squared plus two N plus two over N squared up to plus two N plus N over N squared. So we can factor out a one over and squared um times well two N plus one plus two. N plus two. Up to plus two N percent. Mm. Okay? Um which is equal to what was equal to one over and squared times the sum. Um times the sum where we have I going from one to end of to end and then plus one over N squared times the sum where we have I going from one to end of just I that's equal to while one over N square times to N squared plus one over n square times and times and plus 1/2. Which gives us two plus well and squared over two N squared plus n over two and squared which is equal to what we just get a two and it's just a plus one half. Right and square and square is just one. This is the two plus one half and then end over two and square. Just one over to end. So plus 1/2. And now we can let end 10 off to infinity and end tens off to infinity. One over to end goes to zero. And in the denominator that term goes to zero and we see the area becomes just two plus one half. Therefore we solved the area here is going to be equal to or two plus one half is 2.5. Um So we solved here for the area under the curve and this is equivalent to the distance travel. So therefore we have that are distance traveled is going to be equal to 2.5 ft. Yeah. All right. Take care. Yeah.

Using the following definition. We can find the velocity. Said Avila City is the derivative of our distance, which is D. S. Over the T. And this is equal to one divided by its aim. You have 40 cube minus 42 minus 42 C squared Plus 120 T. This is giving us T divided by sane. You have four C squared -42 t. Blood 1 20 And this is equal to T divided by five. You have two. See squared That is if you divide by two minutes, This is 42, 42. You have Fortitude Cities. Yes, so I have 21. See clad 60. Then for acceleration acceleration A. C. It's equal to The derivative of velocity. And this will give us one divided by five. You have six C squared minus 42. See lads 16. So this acceleration at saru If a celebration equal to zero implies. All right, So we are saying our celebration A. Of T. It's equal to zero. So then it implies that's 60 squared -42 teeth plus 60 will be equal to. So you can divide by six. If you divide by six you have t squared minus 70 Plus 10 to be equal to zero. So to find the roots of this quadratic equation we use the quadratic formula. So do you find that? And if you do you realize that's your C. So we have yeah we have days to be equal to mine. Its servants squared minus four one c. is 10. And this is equal to nine. So it implies that the roots of dates is going to be go to three. So what it means is that T1 is going to be 7 -3 divided by two which is equal to two 82 will be equal to seven plus three divided by two. And this is equal to five. So at acceleration you called zero means that It's a sad c equal to two. So I'd see equal to 20 C equal to five acceleration. It's equal to zero. So to find the velocity at this time we have V At time two is going to be called to one divided by five. You have to time soon Q -21 times two squared Last 60 times two. And this is equal to saying 52. This is equal to 52 Divided by five which is equal to 10. Right? It's full. Then finding the velocity V at a time five is equal to one divided by five. You have to five Q -21. Five squared plus 60 times five and this is equal to five. So then our final results will be fee at that time to its equal to 10 points for nvs five Time we called to 5 to be called to five

Mhm. So for this problem we have a particle moving along an X axis with constant acceleration has velocity of three ft per second at time T one and velocity of negative 1ft per second feet per second at time equals four. So we want to find um the acceleration of the particle so knowing that there's constant acceleration that tells us that um the slope of the velocity graph is acceleration. So if we take the tea or the V two minus B one And divide that whole thing by T 2 -11 will end up with our acceleration. So we see that we start off with or we end with a negative one velocity. And we started off with a positive three velocity And then the time went from 1-4. So we have 4 -1. All right there give us a negative 1.3 repeating. Which tells us that um are acceleration since this is -1 -3 And this is for -1. We end up having a negative for 3m or feet protecting squared acceleration. And we want to find an equation that expresses V as a function of T. So we'll end up with, in this case y equals A to 4/3 t. or in this case the acceleration, did you go to that or the velocity rather is equal to that and then plus B. And we find that why intercept? By plugging in a value? Um And when time equals zero, that will give us um Whatever Y intercept is that will be our value.

And this problem. We're told that starting from Mexico zero with no initial will last. The particle is given an acceleration given by this expression here A equals 0.1 kind of squared of feet squared plus 16 feet per second squared. And we want to determine the position of the particle when V equals three feet per second. Well, we can derive our relationship between velocity as a function of e. Our acceleration as a function of V and acceleration is D v v X t x t t. So DVD X times V and then we can provide this up and in a great move this over here and we get an integral of the X on one side and then V over a TV on the other side. So we called that initially at X equals zero. We have a velocity of zero, and we want to know what the position is. When people ask these three feet per second so we can do the integral, we get over here simply X, and over here we get 10 times square root off 16 plus B squared, evaluated between zero and three. And if we do this evaluation, we get that the position is 10 feet. Now we're asked the speed and the ex elevation of the particle when X equals four feet. Well, when X is four feet, we have he can can use are integral. But now we have a definite integral Over here, we're going from 0 to 4 feet, and over here we're gonna be going from zero to be We can do these inter grows and we wind up with an equation that says 4.4 equals square root of 16 1st three b squared and we can solve for VB, and we get plus or minus 1.83 feet per second. Now, since displacement waas 10 feet when the displacement was 10 feet, the velocity was three feet per second. We're probably gonna We're gonna take the first value here, and we're along that branch because when the position is four feet, we expect to have a positive velocity. So there's two branches here, but here, we're gonna we're on the positive branch. And since we know the velocity at for feet, we can just plug that into our expression for the acceleration in terms of the velocity since this is a last year. Four feet I'm and then we just get 0.44 feet per second squared.


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