The position of a particle as a function of time is given by 𝓇 = ( 5.0î ＋4.0ĵ )t² m where t is in seconds. a. What is the particle's distance from the origin at t = 0, 2, and 5 s?

Video transcript

Hey, everyone, this problem is dealing with vectors. Let's see what they're asking us. We know that an object starts from the origin, were asked to determine its position how it varies with time at two different times. Um Part one is T equals three seconds of part two is T equals eight seconds. The position changes over time according to this given function where three and two are constants and the time is measured in seconds. So we're gonna write this out three I plus two J R R constants are variable T squared and we're working in the units of meters. So the first thing we're going to do is multiply through that variable. So it's gonna look like three T squared in the I direction plus two T squared in the J direction. And from here, we can recall that the magnitude of a vector like this is given as the square root of the sum of squares. So the magnitude of our is going to be three T squared squared plus two T squared squared squared that we come up with the square root of 13 T to the fourth. And we plug that in, we get 3.61 times T squared. So that's our R of T. So at a position at a time, T, our position is going to follow this function. So from here, we can just plug in the two numbers for each of these two parts of the problem. So for part one, uh it was T equals three seconds. So that looks like our Equals 3.6, 1 times three squared And that equals 32. m. And for part two T equals eight seconds, We're going to do the same thing. R equals 3.61 times eight squared. And that is 231 m. So that's the answer to both parts of this problem. We go back to our potential choices that aligns with choice. A so A is the correct answer for this one. That's all we have for this problem. We'll see you in the next video.