A particle's trajectory is described by x = (1/2t^2 - 2t^2)m and ...

Question

A particle's trajectory is described by x = (1/2t^2 - 2t^2)m and y = (1/2 t^2-2t)m, where t is in s. (a) What are the particle's position and speed at t = 0 s and t = 4s?

Accepted Answer

Hey, everyone in this problem, we're told to suppose that a particle's position depends on time and it moves according to the equation X is equal to one third T cubed minus T squared meters and Y is equal to 1/4 T squared minus T meters. Whereas to determine where the particle is at T equals zero seconds and T equals six seconds. Okay. And also find its speed at the given time intervals. Alright. So let's start with part one. Okay. We have T is equal to zero seconds and the position of our particle are, is going to be given by the X coordinate and the Y coordinate. So let's start with the X coordinate. We know that X is equal to one third T cubed minus T squared meters. Okay. This is the exposition given in the problem. If we substitute T equals zero seconds, here we get that X is equal to zero m. And then similarly for Y, the Y position in the problem is given by Y is equal to 1/ T squared minus T in meters. And again, if we substitute T equals zero, we get zero minus zero. We just get zero m. And so the position of our particle at T equals zero seconds, we're gonna call it R zero is equal to zero. I had K zero in the X direction. Recalled that the I hat represents the unit vector in the X direction plus zero J hat K J hat similarly represents a unit vector in the Y direction. Meters. Alright, so that's our position vector for T equals zero seconds. If we look at our answer traces, we can see that we can already eliminate answer choice C because it has a position vector at T equals zero seconds of two I plus two J hot meters, which is not what we have. So we can eliminate C already. Now, before we move on to the second time point, let's do the speed. Okay. We're asked to calculate the speed as well. Now when we're talking about speed, recall that the speed is related to the position through the derivative. Okay. So the speed V is equal to the derivative of the position with respect to time. So we can take the derivative of each of these components, the X component and the white component we have D X by D T. And I'm gonna call this V X. This is gonna be the X component of our speed or velocity. We take the derivative of one third T cubed minus T squared T cubed, we bring the three down. Okay. We have 3/3 which become one T squared. So we get T squared and then the derivative of T squared is two T. So we have T squared minus two T. Our unit here is now meter per second because we were brought one of these teas down. Okay. And if we substitute T equals zero seconds, we get that. This is equal to zero. It's not zero seconds, zero m per second. Okay. T squared minus two, T one T is zero is just zero. We can do the same for the Y component. V Y is D Y by D T. Where, why is the position? Now we have 1/4 T squared minus T the derivative here is gonna be one half T minus one meters per second. And when we substitute T equals zero, we get zero minus one m per second. And so are Y velocity is negative one m per second. So we can write our velocity V just like we wrote the position and I'm gonna put V not to indicate V at zero seconds and this is equal to zero. I hat. Okay. We had zero in the X component plus negative one. J had negative one in the Y component and the unit is meters per second. Now we're asked to find the speed, not the velocity vector. So if we look at the answer choices, we see that we want the magnitude of the speed and so the magnitude of the speed, the not okay. It's gonna be the magnitude of this vector. Recall that to find the magnitude of a vector, we have the square root some of the two components squirt. So we have the square root of zero m per second squared plus negative one m per second squared, just going to be equal to the square root of one m squared per second squared. And when we take the square root, we just get one m per second. So for time to equal zero seconds, we found the position vector zero, I hat plus zero J hat meters when we found the speed of one m per second. Now let's move on to part two where we have the time is six seconds, we're going to do the exact same process. We're just going to substitute in a different time point. So for T equals six seconds, we get that X is equal to and again, our X is one third, Okay. T. Cubed, six seconds, cubed minus T squared, six seconds squirt in meters a six square or six seconds. Cute times 1/3 is going to give us 72 And then we have six seconds squared, Gives us 36 seconds scored. Our unit is meters and so we get 72 - m, which is equal to 36 m. Similarly, in the y direction, We get 1/4 times six squared minus six. The unit is m. Six squared gives us 36 divided by four, gives us nine. So we get nine minus six m, which gives us three m. So if we want to write this, in terms of our position vector R, I'm gonna break our sub six to indicate that we're talking about the position at six seconds. We have 36. I had K 36 in the X direction plus three J hat three in the Y direction in our unit is meters. So that is our position vector. Now, we can move on to our velocity vector. Okay. And finding that speed and the velocity we've already taken the derivative. So we don't need to take the derivative again. And so V X, we're going to use the same equation we used for zero seconds which was T squared minus two T meters per second. So we get T squared -2 T m/s. This is equal to six squared minus two times six meters per second, which gives us a speed or velocity in the X direction of 24 meters per second in the Y direction. Similarly, okay. We found that the Y component of the velocity can be given by one half T minus one m per second. So you get 1/2 meter per second. Substituting in T is equal to six seconds. We have one half times six minus one meter per second, one half times six gives us three minus one gives us two m per second. Alright, now we can write our velocity vector again for six seconds. We have 24 m per second in the X direction. So 24 I had plus two m per second J hat in the Y direction, meter per second. And if we want to find the speed again, we can take the magnitude of this vector which is gonna be equal to the square root of each component squared, 24 squared plus two squared. And if we work this out, and these units are meters per second on 24 2, We get a speed of approximately 24.08 m/s. And that is for that 6 2nd time point. So if we go back up to our answer traces, now we have four quantities that we found. We found that the position vector of the particle at T equals zero seconds with zero I hat plus zero J hot meters. And we found that it had a speed of one m per second. Okay. So A B or D all have those same answers. So we can consider those. Now for the particle at T equals six seconds, we found a position vector of 36 I hat plus three J had we found a speed of approximately 24.1 m per second. And so we have answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

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A particle's trajectory is described by x = (1/2t^2 - 2t^2)m and y = (1/2 t^2-2t)m, where t is in s. (a) What are the particle's position and speed at t = 0 s and t = 4s?