Suppose the position of an object moving horizontally along a ...

Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
Determine the acceleration of the object when its velocity is zero.
f(t) = t² - 4t; 0 ≤ t ≤ 5

Determine the acceleration of the object when its velocity is zero.

f(t) = t² - 4t; 0 ≤ t ≤ 5

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says an object moves along a straight path, and its position after T seconds is described by the function P of T is equal to TQ minus 6 T2 plus 9T, where P is measured in meters. Determine the object's acceleration when its velocity is zero. We're given 4 possible choices as our answers. For choice A, the acceleration is minus 6 m per second squared when T is equal to 1 and 6 m per second squared when T is equal to 3. For choice B, the acceleration is 6 m per second squared when T is equal to 1, and minus 6 m per second squared when T is equal to 3. For choice C, the acceleration is 12 m per second squared when T is equal to 1, and minus 12 m per second squared when T is equal to 3. And for choice D, the acceleration is 16 m per second squared when T is equal to 1, and minus 16 m per second squared when T is equal to 3. Now we're asked to determine the object's acceleration when its velocity is 0. So, in order to determine the veloc when the velocity is zero, we need to define our velocity as a function of time. And recall that our velocity is the time derivative of our position equation. So that means our velocity, BFT. It's going to be equal to the derivative with respect to T. Of PFT since that is our position function. And so this is going to be equal to the derivative with respect to T of the quantity of Ted minus 6 T squared. Plus 90. So we need to take a derivative of this expression that this is going to be equal to. Our first term we'll use our power rule. So we call for the power rule, we're going to multiply our variable by the exponent. So we'll have 3 multiplied by T raised to the quantity of 3 minus 1, since we need to subtract 1 from the exponent. Then we'll have minus. For our second term, we'll take the derivative of 6 T2d. So here we're going to apply our constant multiple and power rule together. So we'll have our constant 6 multiplied by the exponent, which is 2, and that gets multiplied by T raised to the quantity of 2 minus 1. Then we'll have plus. For the last term, we'll take a derivative of 9 T, and again, we'll here apply our constant multiple and power rule. So we'll have 9 multiplied by the exponent, which is 1, multiplied by T raised to the 1 minus 1. So simplifying this expression, this is going to be equal to 3 T squared minus 12 T. Plus 9. So this here is our velocity as a function of time. Now, I need to determine when the velocity is 0, so I'm going to set this expression equal to 0 and solve for T. So here I have a quadratic function, and we can actually factor this quadratic. And so factoring it, we'll get the quantity of 3 T minus 3 in quantity, multiplied by the quantity of T minus 3 in quantity, and this has to be equal to 0. So now what I'm going to do is set each factor to 0 and solve for T. So, for the first factor, we'll have 3 T minus 3 is equal to 0, and then we'll have T is equal to 1. For the second factor, we'll have T minus 3 is equal to 0, and solving for T, we'll get T is equal to 3. So these are the two times that I am looking for. And so I'll need to determine what our acceleration is at these times. So in order to do that, I first need to figure out our acceleration as a function of time. And to recall that our acceleration is equal to the time derivative of our velocity. So that means that AFT, which is our acceleration as a function of time, is going to be equal to the derivative with respect to T of VFT. And since we found an expression for V of T, we'll just substitute that in, so this is going to be equal to the derivative with respect to T of the quantity of 3 T squad minus 12 T. Plus 9. So now I need to take a derivative of this expression. So doing so, this is going to be equal to. For the first term, we're going to use our constant multiple and power rule. So we're gonna have our constant 3 multiplied by the exponent 2, and that gets multiplied by T raised to the quantity of 2 minus 1. Then we'll have minus, we'll take a derivative of our second term, which is 12 T. So again, applying our constant multiple in power rules, we'll have 12 multiplied by 1, multiplied by T raised to the 1 minus 1. And then for the last term, we're taking a derivative of 9 with respect to T. And so here we need to recall our um constant rule for derivatives. So anytime we take a derivative of a constant, it's going to be 0, so we'll have plus 0 as our final term. So simplifying this expression, this is going to be equal to 6 T minus 12. So now we need to just determine our acceleration when T is equal to 1 and T is equal to 3. So, we'll first calculate our acceleration when T is equal to 1. So I'm looking for A of 1. This is going to be equal to 6, multiplied by 1. -12, and this is going to be equal to -6 m per second squared. We'll also need to calculate AF 3. This is going to be equal to 6, multiplied by 3. -12, and this is going to be equal to 6 m per second squared. So those are the two accelerations that we are looking for. And the units of meters per second square come from the fact that our position was given in meters and our time was given in seconds, and we've taken two derivatives of our position function in order to get our acceleration. So that's the reason why we get units of meters per second squared. So now that we have our answers, we can compare our calculated answers to the answer choices, and we see that the correct answer is answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Position, Velocity, and Acceleration

Finding Critical Points

Evaluating Derivatives

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