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Find the velocity vector v(t) given the acceleration vector a(t) and the initial velocity: a(t) (7t, 2) , v(o) (3, 5)v(t)...

Question

Find the velocity vector v(t) given the acceleration vector a(t) and the initial velocity: a(t) (7t, 2) , v(o) (3, 5)v(t)

Find the velocity vector v(t) given the acceleration vector a(t) and the initial velocity: a(t) (7t, 2) , v(o) (3, 5) v(t)



Answers

Find the velocity and acceleration vectors. $$x=t, y=t^{3}-t$$

Okay, so for this exercise we have the following equations that represent some motion vector. So the motion picture following this equation is going to be represented by T. T square. And take you. Now we need to find the velocity vector and the acceleration vector. You can also write this in a different location. That could be T. I lost t square. J. Lost t Q K. Okay. So now we need to calculate the velocity vector and to find the velocity vector, we need to take the further relative of the motion vector. So the derivative of T is going to be one. Lastly the relative of T square to T plus the derivative of the cube. That's going to be three T square. Okay then we need the acceleration vector and for the acceleration acceleration vector we need to take the derivative the first derivative of the velocity vector or the second derivative of the motion factory, which is the same. So in this case the second, the relative of one, it's going to be zero. The relative of duty, it's going to be too And the derivative of three T square is going to be just 60. Okay, so at the end the acceleration vector, it's going to be two Day Plus six T. Okay,

Hi there. So for this exercise we have here equations that represents a linear motion. Let's remember that a linear motion is described by a objective function that starts from an initial point. That is a vector plus some function that depends on T times some velocity vector. Okay, that gives you the direction and how this is going to evolve linearly. That's why this represents a linear motion. Okay, so this is the general structure and you can observe that these equations here in particular are given in that sense because we can write the equation of mood, the vector motion of this particle As the initial .235 plus T square. And the velocity vector that in this case is going to be one -2 and -1. Okay, so it's a linear motion. Now, what we are interested for this exercise is to obtain the velocity picture and the acceleration vector of this motion. And then we need to check what happened if we need to check actually that the acceleration and the velocity are going to be a line. If the velocity increase and in case the velocity start to decrease then that means that the acceleration and velocity are pointing to different directions, two opposite directions to be more precise. So let's calculate what's going to be the velocity vector for this motion. So the velocity vector is equals to the first derivative of the motion picture. This part is going to be constant. So we're not going to take into account and also forgot forgot um to grade this in different notations. So it is going to be equal to um two plus t squared I Plus 3 -2 T sq jay and plus five minus t K. Okay, so the first relative it's going to be equal to tee times I minus T j minus. We just need to take the derivative of this part so that the relativist to T. Yes. So this is going to be the velocity vector now, the acceleration vector. We need to debate again the velocity factory And that is equal to two times I -2 J minus K. Or what is equivalent to I- for day -2. Okay, okay, so these are the velocity and acceleration. Now we need to check what happened if we start to increase the speed of this particle or to decrease this bit of this particle. So you can observe hear that if we take well the speeds is obtained by taking the norm of the velocity vector and that is going to be equal to two t times the square root of one plus four plus one. What is equivalent To T. Times the Square Root of six. Okay, so what happened here? The increasing the velocity is a fine for So let me put it down here, I'm going to know days by just normal b. Okay, so we increase For T greater than zero and if t is starting to go for for for low with values Okay, yes to to to negative values then the the velocity is going to decrease. Yeah. Okay, so The increase 40 greater than zero. And the decrease for T less than zero. Now, how we can check if these two vectors are aligned in this case. Because what we need to show is that if the velocity increase then the the director, the velocity vector and the acceleration vector are pointed to the same direction and that can be obtained if the dot product of this too. Our is positive for T greater than zero. So these are product is going to be equal two, two. Okay. To fourty times one plus four plus one. Okay, this is equal to 24 T. And you can observe that if T is greater than zero than this value is positive. So that means that these two vectors are aligned. Yeah. Now what happens if the dot product is uh negative? Well, if the product is negative, that means that they are pointed in different directions. And we know that these values are close to 2014 and this clearly is negative for uh for values of less than zero. So that means that in this case the acceleration is pointing to the opposite direction uh BB

Uh huh. In this problem we wish to find the velocity V. And acceleration a of an object whose distance asked from the origin at time. T is S M T equals 4.90 square plus five to this question is challenging understanding implicitly of differentiation. The reason that is the case is because in rectilinear motion the velocity V is the river I have asked with respect to T D S D T. The acceleration is also displayed STT square or D V D C T. Acceleration is also the second derivative of S. This question is also testing our understanding of higher or determines where D N Y D X. And is the end of the Y found by taking and the reason why successively so to solve, we're going to utilize the above information in conjunction with previously learned differentiation shortcuts to solve. Thus we have vehicles. Ds DT equals 4.19 to 30 plus five or 9.80 plus five as shown Were 9.1804 point and he squared 515. Thus, by the power rule and constant role A is simply 9.80 within 3- 50 or simply a is 9.8.

In discussion. We have the position vector of a particle that is given by are the vehicles to three T square coma to take you? We are required to find develop city vector and the exploration vector for this position vector. So let's see how this whole discussion we know that develop native actor VT can be readiness the first derivative or D. By duty of position vector. Artie. So on the behavior of the formula we can right Velocity vector we t. is equals two D by duty. And the position victories three T square coma two D cube. So this will be close to be by your duty three T square coma D by duty to pay cube. So when we differentiate these terms we get the velocity vector that is The differentiation of 30 square will be called to two into three. There have been six T comma the differentiation of to tick you will be called to six piece choir. Hence we can conclude that D velocity vector is 60 Coma six T square. Right, so this is a final answer for the part A. And now let's move to part B. Okay, part B. And in this part of where we are required to find the acceleration vector. The formula for the acceleration vector, 80 can grittiness D by duty of velocity vector. Bt We have already calculated the velocity vector so we can write D by duty of 60 coma six t square. Hence this will be called to D by duty of 60 coma D by duty of six T square. So then we differentiated we get declaration, 80 is recalled to. The differentiation of 60 will be called to six comma. The differentiation of sixties who will be called to 12 P. Hence you can conclude that the acceleration vector is six comma 20. So this is a final answer for this problem. I hope you understood the solution. Thank you.


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