A particle moving along the x-axis has its veocity described by t...

Question

A particle moving along the x-axis has its veocity described by the function vx = 2t^2 m/s, where t is in s. itsinitial position is x0 = 1 m at t0 = 0 s. At t = 1 s what are the particle's (a) position

Accepted Answer

Hey, everyone in this problem, the velocity of a shred traveling in the Y direction follows V Y is equal to three T squared meters per second. For the time is in seconds at time. T not equal zero seconds. The starting position of the shred is why not equals two m were asked to calculate the position of the particle at T equals one second. The answer choices were given our A equals three m. B equals one m C equals two m and D equals zero m. Now, we have to be careful with a problem like this. Okay. It's very tempting to go ahead and find the not using this V Y equation. We can find V at one second which would be our final V and then use our Kinnah Matic or you am equations. However, notice I recall that the acceleration in this case, in the Y direction is equal to the derivative of V Y with respect to D T. And in this case, V Y is three T squared. If we were to take the derivative, we would get six T and this is not a constant, not equal to a constant. What this means is we don't have uniform acceleration. So we can't use our, you am uniformly accelerated motion equations. So what do we do? Instead, we'll recall that we can write the change in position delta Y as the integral from T equals T, not two T equals t final our velocity in the Y direction B Y. Yeah, that's a function of T and we take the integral with respect to T. Alright. What is delta Y delta Y is our final Y value minus our initial Y value In that final y value the position of the particle at 1/2. That's what we're looking for. This is gonna equal the integral from T equals, well, it's our initial time 0.0 seconds starting point okay in the position or the time that we're looking for the particular position is one second and final time is going to be one second. So we have the integral from zero seconds to one seconds of this V Y which is three T squared meters per second D T. Alright. So now we just need to take that integral. And what about why not? We're told that at time T equals zero seconds. The starting position of the shred is that why not equals two neighbors? So we have Y F minus two m is equal to if we take the integral of three T squared and we raise the power of the exponents. So we get T Cute and then we divide by that new expo. So we have 3/3 T cubed. The unit here is going to be meters and we're going to evaluate from T equals zero seconds to T equals one second. Now, I just want to make a note on the unit before we had three T squared. And when we talk about units, I'm gonna put this in square brackets. We know that the units were meters per second. Okay. So square brackets just to represent units T squared has a unit of seconds squared and this is equal to meters per second. That means that this value of three must have had a unit of meters per second cubed. Okay. So that constant three must have had a unit three or sorry meters per second. Cute in order for three T squared to have the unit of meters per second. So when we take the integral, we still have that constant meters per second cubed. And this is after we take the integral. So I'm gonna draw a line and now we have a T cubed value. So now we have seconds cubed instead of second squared, the second cube units cancel and we're left with just a unit of meters. Alright, perfect. Let's carry on with our calculations. We have Y F minus two m is equal to 3/3, gives us just one T cubed meters evaluated one second. So we have one cube meters minus the evaluation at our initial time point which is 00 cubic meters. This is just gonna give us one m and we can move the two m over to the right hand side to isolate Y F. We get Y F is equal to one m plus two m which is going to give us a final position of three m. We go back up to our answer traces and compare what we found. We see that we have answer choice. A, the position of the particle at T equals one second is three m. Thanks everyone for watching. I hope this video helped you in the next one.

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A particle moving along the x-axis has its veocity described by the function vx = 2t^2 m/s, where t is in s. itsinitial position is x0 = 1 m at t0 = 0 s. At t = 1 s what are the particle's (a) position

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