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A particle is moving in space with a distance function s(t) %ta 2t2 + 3t melers in t seconds. a. Determine the time in seconds when the particle is at rest: b. Eval...

Question

A particle is moving in space with a distance function s(t) %ta 2t2 + 3t melers in t seconds. a. Determine the time in seconds when the particle is at rest: b. Evaluate the acceleration of the particle when the velocity is 0. (10 marks)Differentiate f (x) = (x + 1)3(18x+ + (simplify your answer as much as possible)(5 Marks)Find f (0) and f" (x) for f (x)

A particle is moving in space with a distance function s(t) %ta 2t2 + 3t melers in t seconds. a. Determine the time in seconds when the particle is at rest: b. Evaluate the acceleration of the particle when the velocity is 0. (10 marks) Differentiate f (x) = (x + 1)3(18x+ + (simplify your answer as much as possible) (5 Marks) Find f (0) and f" (x) for f (x)



Answers

A particle moves along a straight line with equation of motion $ s = f(t) $, where $ s $ is measured in meters and $ t $ in seconds. Find the velocity and the speed when $ t = 4 $.

$ f(t) = 10 + \frac{45}{t + 1} $

In this problem, we are beginning to get comfortable with taking first derivatives. Specifically, we are now making the connection between derivatives and its applications specifically with finding velocity. So we are given a function f of t and f of T equals 10 plus 45 all over t plus one and we're told find the velocity given that this equation is the position. Now, what do we know about the relationship between position and velocity? Velocity is the first derivative of position. So what? This problem is essentially asking you is to simply take the first derivative of F 50. So we have to find f prime of tea. Now, the first thing to note is that the derivative of a constant is zero. And here we have a quotient. So we're going to be using quotient rule. So f prime of T is going to be equal to negative 45 times the derivative of T plus the derivative of one all over T plus one squared. And we're going to simplify this to get negative 45 times one plus zero all over T plus one squared. So then we can simplify even further to get F prime of T equals negative 45 all over T plus one squared, and that is the equation for velocity. However, that's not the end of this problem, we're told. What's the velocity of the particle moving at a specific time, which we call t and what is what is the velocity when t equals for So all we have to do is plug in four 40 into our derivative F prime of four would be equal to negative 45 all over four plus one squared. So that's equal to negative 45 all over 25 which simplifies to negative 9/5. So what we can say is that our velocity at time four via four equals negative 9/5 meters per second. So I hope that this problem helped you understand a little bit more about the relationship between position and velocity and how we can understand our relationship using the concept of the derivative

Or go for this problem. Um And um for the following problem is to consider the position function given to us, which in this case is S. F. T equals t square. I have three peoples too. And with this model we can find so much information. We see that um we can find things such as instantaneous velocity, wow, we cannot find average velocity. We want to find displacement over a certain amount of time, such as five seconds. We can plug that in there. Um We can also consider um average velocity which is going to be S f E minus s of a all divided by being my inside. This can be average velocity. If we want to look at graphs for example and see when it reaches at the highest point, um we see that this right here is going to be um something that is going to model a particle moving along a line with its position. So it's going to look different than, say if we had a rocket that had parabolic trajectory where it reaches a maximum point. If we have something like that, we're able to see where that maximum point is reached, where it hits the ground and that's gonna be helpful in solving other problems.

Hello Today we're going to solve a simple, simple integration problem. So let's get started. So an acceleration function, we're going to define it as eight times t It's going to be defined s t plus four where t is the time. So what we're gonna do is first we are going to find the velocity at time t So the the last city at time t at Time T why is it he? Well, it means that it's an arbitrary type. So given any amount of time, right, we're gonna we're gonna have a function where we're going to tell what the velocity is in respect to that point in time. So we're not giving an exact amount, which is giving out a formula. And then we is going to be the distant travel. Oh, oops. Travel during a time interval from zero to 10, all inclusive. So from 0 to 10 and let me set the initial condition. At T equals zero V equals five. So let's get started. So let's first refresh our knowledge. What is acceleration? Acceleration is the rate of change of velocity in respect. Two time, right, So that's going to be dealt with he I mean Delta V over Delta t. Well, in the case of calculus, it's going to be an instantaneous change, meaning it's going to be the derivative of V in respect to t and so therefore, manipulating dysfunction. What we're going to have is that when we multiply both sides by dft, we're going to have that right. So we're going to have d V equals a times d f t. If we were to integrate this, then the differential and the interval they cancel out we're going to have V as a function of t equals two, the anti derivative of a times d of t. So basically, to find the velocity function, all we have to do is integrate the acceleration function. So the integral of a f T in respect two d f t is going to be the integral of T plus four times DT. So using the power role, we can conclude that the anti derivative of T is going to be t squared, divided by two plus 40 plus a constant C and that is going to be our velocity function. There it is. So how do we solve for this constant? See here. Well, remember that we had this initial property. So what we can do is at every point of t we see we're gonna plug in to equal zero, and then the function is going to output a five. So when t is zero, then the velocity is going to be five. So five equals zero square, divided by two plus four times zero equals Steve. So zero squared 00 developer to zero. Four times zero is zero. And so therefore, C is equal to five. And now we can substitute to see 25 and there we have it. This is our velocity function. So whenever we have t, we can just plug it into the function and we're going to get the velocity. So that's the first part. Now, the second part, we're going to find the distance that we have traveled. Okay, So how do we do this? Well, first of all, we have to figure out does this function have any negative value at any time? So let's take a look. This function the first coefficient is positive. Second is positive and it's being shifted up by five, which means that it's going to probably look something like this, so we don't have to really worry about anything that's down there, because if it's down there, then there's going to be a negative and the distance is the absolute value. But since we don't have to worry about that, all we have to do now is just integrate from one to the other limit of integration. All right, so first of all, how do we define our distance? Well, remember that the distance is the derivative. I mean, the integral of the in respect to t to d. T. So that means the distant function is going to be dft time's d t from 0 to 10 inclusive, so 0 to 10. So that's going to be the same as T squared by two plus 40 plus five time's d T, which we will get T Cube divided by two times three. That's gonna be six plus to t squared. Oops, I'm sorry. Why they put that there. This is our anti derivative already plus five t plus the right. But now, because we're calculating the definite integral, we don't really have to care about the sea because it's going to get cancelled out anyways. So let's check if this is right or not. Using the power roll three. We bring it down three and six to cancel out. We get to that's correct. And then to we bring it down, get 45 We take out the T, we get five. Perfect. So now all we have to do is plug in our limits of integration. So first we have 1000. What about six plus 200 plus 50 and then the lower bound is going to be zero. And since all of them are T, then that whole thing is going to evaluate to zero. So we don't have to worry about that. So this will equal to 416 67 inches and that's it. That's your answer. If we were to be told to run it up, it's gonna be 4 67 inches. That's it. There you go.

So we're given the function of acceleration to be T. Is equal to you. D plus four. Were also given that the velocity is equal to five at time zero. As well as the interval of time being 0-10 seconds. So for part we want to find the velocity at time T in order to find the velocity with are given information. The first thing we want to do is find the integral of our acceleration function. The integral of acceleration will give us velocity. So we have the integral uh T plus four DT. This will end up giving us t squared over two plus 14. Quite see. Can't forget your plus C. Mhm. So next now that we know that the velocity is equal to five at time zero. We're going to plug in zero for everywhere where we see A. T. So we have 0/2 Plus four times 0 plus C. At time zero. So this will end up equaling see and since we know that uh time zero philosophy, 0 to 5 C will end up equaling five. So now from here we can get our velocity with respect to time function. The equal t squared over two plus 14 plus five. And this will be our function for velocity. So next up for part B we want to find the distance traveled during the time integral time interval. So in order to do that first we're going to want to take the integral uh Our time interval 0-10 with respect to our velocity function. The integral of velocity is distance. So B. F T D T. So we're simply going to plug in our velocity function here into our integral. So we have 0-10 of t squared over two plus 40 plus five. Yeah. Okay, so first step will be to evaluate the integral and once we evaluate our integral you should get Teak You'd ever six plus to t squared plus five T. And we want to evaluate this integral once again on the time interval from 0 to 10. So since we have 0-10, we're going to plug in 10 everywhere where we see our T. And we can ignore the second half. We're plugging in zero because when we plug in zero, the whole function is going to equal zero. So you end up with 10 cubed over six plus two times 10 squared Plus five times 10. And when you evaluate this out, You should have 1000 over six plus 200 plus 50. When you evaluate this out, finally, your final answer will end up being about 416 0.67. And this will be the distance that was traveled.


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