The position of a particle is given by the function x = (2t^3 = 6...

Question

The position of a particle is given by the function x = (2t^3 = 6t^2 + 12) m, where t is in s. (a) At what time does the particle reach its minimum velocity? What is (vx)min?

Accepted Answer

Hey, everyone in this problem, the position of a water droplet cruising in the air follows the function Y is equal to T cubed minus eight, multiplied by T squared plus four. And it's given in meters where time is measured in seconds. We're asked to determine the time when that particle has the least philosophy and the value of the least velocity. We're looking for the time and the value we're given four answer choices. Answer choice A has a time of 2.67 seconds And at least velocity of 21.3 m/s. Answer choice B has a time of 2.67 seconds and a least velocity of 0.2 m per second. Answice Choice C has a time of 2.67 seconds and a least velocity of negative 21.3 m per second. And answer choice D has a time of 0. seconds and a least velocity of 2.93 m per second. Now, we're given the position of the water droplet, Y is equal to T Q minus eight multiplied by T squared Plus four. And this is given in m. We're looking for the least velocity in the time when it occurs. So recall that the velocity is related to the position through the derivative. So the velocity is the derivative of the position with respect to time D Y divided by D T. If we take the derivative of the position vector, this is a polynomial. So we have T cube, the exponent is gonna come in front and then lower by one. So we have three T squared and we have minus eight T squared. Again, the exponent of two is gonna come down in front and multiply. And then the exponent is gonna lower by one to minus 16 multiplied by T and the derivative of a constant four is going to be zero. So we have three T squared minus 16 T and the units for velocity are going to be meters per second. So now we have our velocity, we wanna figure out when it is the minimum. OK. So at what time T do we have a minimum value? I wanna note here that our velocity which is equal to three multiplied by T squared -16, multiplied by T is a quadratic function and it's going to open upwards. So let's go ahead and find the critical points. We want to find the minimum. So let's start by finding the critical points and recall that the critical points are going to occur on the derivative of our function of interest. So D V by D T Is equal to zero. Now, if we take the derivative of our velocity with respect to time D V divided by D T three multiplied by T squared, taking the derivative, the exponent of two will come down and multiply in front, the exponent gets lowered by one. So we have six T and we have negative 16, multiplied by T. Again, the T has an exponent of one. So that one comes out in front and multiplies we lower the exponent by one. And so we get T to the exponent zero. And so we have just minus 16 there, We wanna set this equal to zero and we can solve for the T value that will give us this zero value. We have six T is equal to 16, which tells us that the time T is equal to 16 divided by six Is approximately 2.66 repeated seconds. OK. So this is our only critical point. This is where the derivative of our velocity is zero if the derivative is zero, that means that the slope is zero. So if we can think of a parabola opening upwards and we have a slope of zero that's gonna occur at the minimum. And so this critical point, this T value we found is going to be the time Where we get the least velocity. And so we have this t value of 2.66 repeated seconds If we go down and work through part two. Now, we know the T value, we wanna find what that least velocity is. So what we're looking for is the velocity at that particular value 2.66 repeated seconds. This is going to be equal to three, Multiplied by 2.6, 6 repeated second squared minus 16, Multiplied by 2.6, 6 repeated seconds. And all of this is given in meters per second. I recall from that very beginning where we were given the position of meters. We took the derivative and said that it was in meters per second. So all of this is going to give us meters per second. We get 21.3, 3 repeated -42. repeated meters per second For a least velocity of -21. m/s. That is a velocity, the least velocity that we have. So we've done part one and part two. Let's go compare to our answer choices. We found a time of 2.66 repeated seconds. Our answer choice is round to three significant digits. So we're gonna have 2.67 seconds and we found a least velocity of negative 21.3 m per second. So we have answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

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The position of a particle is given by the function x = (2t^3 = 6t^2 + 12) m, where t is in s. (a) At what time does the particle reach its minimum velocity? What is (vx)min?

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