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An object moves along the T-axis with acceleration a(t) 2t 2 units per second per second_ Its position at time t = 0 is 5 units to the right of the origin One secon...

Question

An object moves along the T-axis with acceleration a(t) 2t 2 units per second per second_ Its position at time t = 0 is 5 units to the right of the origin One second Iater the object iS moving Ieft at the rate of 4 units per second.(a) Find the position of the object at timeseconds(D) How far does the object travel during these 4 seconds?

An object moves along the T-axis with acceleration a(t) 2t 2 units per second per second_ Its position at time t = 0 is 5 units to the right of the origin One second Iater the object iS moving Ieft at the rate of 4 units per second. (a) Find the position of the object at time seconds (D) How far does the object travel during these 4 seconds?



Answers

At time $t=0$ a car traveling $25 \mathrm{~m} / \mathrm{s}$ begins to accelerate at a constant rate of $-4 \mathrm{~m} / \mathrm{s}^{2}$. After how many seconds does the car come to a stop and how far will the car have traveled between $t=0$ and the time it stopped?

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.

So here we need to calculate the distance traveled during the excellence. The distance can be returned as questions as This will be equals two. U. T. Less of it is square. Let us substitute the values so you're given you has given us four and time he's equals two fives and plus. Uh huh Multiplied by the acceleration is given as 1.2 m person and square Well played. Where time has given us five seconds. After calculation, we will get the required distance table in the given time. And trouble is Comes out to be 35. Make this so this is the required distance table then we have in Israel Spirit It's a little bit you is equals two four metre purse again And time is equals two. What was the game acceleration during this time? And trouble is it close to 1.2 Metre position is So this we have got us 35 m

In this problem, we will cover the velocity of a function. So to solve the first part of this problem, we know that we want to find the derivative when T. of zero is equal to one. So I'm going to write F prime or not every time S prime of one and we know that it's going to equal the limit. S. T Approaches one of S and t Minour as of one over c minus one. And we know what S. And P. Is. And when we plug in one for S and T, we will get to so we have the square root of four T -2 Over T -1. And in order to get rid of that square root on the top, we will multiply the top and the bottom By the Square root of 14 plus two. And what that gives us is limit as T approaches one, So we get 40 -4 On top, And then T -1 Times the square root of 40 plus two on the bottom. And we see that we could factor the top. So let me write the limit as he approaches one. We can pull out four in front and right T -1, And this is going to be over T -1 Times the square root of 4ty plus two, And we can cancel out the T -1. So we will be left with the limit as t approaches one of four over the square root of 14 plus two, and if we were to plug in one for thi We would be left with four over four, which ultimately becomes not to think one and that is our velocity. And so for the second part Of the problem we will do the same thing. We want to find the derivative when T0 is equal to four. So I'll write as prime four equals the limit. As T approaches for Of S. A. G- S. A. four. All over t minus four. And we already know what S. And P. Is. And when we plug in four for tea we will get four. So the top should be the square root of 4ty minus four Over T -4. And to rationalize this will multiply the job and the bottom by the square root of 40 plus four, same for the bottom. And that yields us the limit as T approaches four of 40 minus 16. This is going to be over t minus four Times the square root of 4ty plus four. And we see that we can factor the top. We can take out of four from both 40 and 16, So we will have four times t minus four All over to you -4 Times The Square Root 40 Plus four. These cancel out, and we are left with the limit as T approaches for of four over the square root of 40 plus four. And when we plug in 440 we will get that are derivatives is going to be four over eight, which can be simplified to one half.

In this problem, we will cover the velocity of a function. So we are given a function F of T. Meant to calculate s the distance of the object based on a variable T, which represents time in seconds. And we want to find the velocity at three different times. So the first is when the initial time is equal to zero and we know that the velocity is the derivative. So it has the form of The limit as the change in T approaches zero of the change in s over the change in T. And we can rewrite this as the limit As T approaches our initial time, which is zero. Uh F f t minus F of zero Over G zero. And we know F and G of course is three T squared plus 40 & F of zero, we can see that it's going to be zero, so nothing to add there in the bottom is going to be T. Yeah. And we see that that she can be factored out of both terms. So let me write the limit As she approaches zero. And since a. T. Can be factored out of Both terms, we get the simplified version which is three T plus four. and when we plug in T0 we get that the limit Would be equal to four. So that's our first velocity and it's in meters per second. So now our initial time is going to be too. And so once again we have our velocity. It's going to be equal to the limit. S. T. Approaches to this time and that's A F. G minus f. Of two over t minus two. And again we have three T squared plus four teeth -F two. And this fft was going to be 20 So we have -20 Over G -2. So now we have the limit she approaches to and we can factor out the top so we get t minus two, multiplied by free t last time over T -2. We see that that T -2 from the bottom can be cancelled out with the T -2 from the top. So we are left with the limit S. T approaches to Of three T. Question. And when we plug in two for tea we get that our velocity is going to be 16 m/s last. We want to find the velocity at any initial time. So now we have the equals to limit As T approaches T0 F of G -F. Over G -G zero. That's going to give us on top three, T squared plus 40 minus three t zero squared minus four t zero over t minus t zero. If we were to factor out top We would get T -G0 times three T plus three cheese zero plus four. All over T zero. Of course, those t minus t zeros cancel out. And we are going to be left with The limit as she approaches T0 of three T plus three T zero plus four. and when we plug in T0 for tea, we will get color 60 0 plus four meters per second as our velocity.


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