a-particle-moves-according-to-a-law-of-motion-s-ft-t-ge-0-where-t-is-measured-in-seconds-and-s-in-4

Question

A particle moves according to a law of motion $ s = f(t), t \ge 0, $ where $ t $ is measured in seconds and $ s $ in feet.
(a) Find the velocity at time $ t. $
(b) What is the velocity after I second? 
(c) When is the particle at rest?
(d) When is the particle moving in the positive direction?
(e) Find the total distance traveled during the first 6 seconds.
(f) Draw a diagram like Figure 2 to illustrate the motion of the particle.
(g) Find the acceleration at time $ t $ and after 1 second.
(h) Graph the position, velocity, and acceleration functions for $ 0 \le t \le 6. $
(i) When is the particle speeding up? When is it slowing down?

$ f(t) = t^2 e^{-t} $

Heather Zimmers · Accepted Answer

here we have a position equation and for part A, we want to find the velocity. Remember that velocity is the derivative of position. So we&#x27;re going to use the product rule to find the velocity. So we have the first times a derivative of the second plus the second time&#x27;s a derivative of the first, and we had to use a chain rule to find the derivative of each of the negative T. Now we want to simplify that. So what I&#x27;ve done is noticed that you can factor and e to the negative t out of both terms. And you can also factor a t out of both terms. So I did that and we get tee times each of the negative t multiplied by negative T plus two. And then to further simplify, I just changed the third term for the third factor. And I called it t minus are excuse me to minus T instead of negative T plus two. So that gets us in a good form where we can use it more easily when we move on for Part B were finding the of one. So we substitute one into that velocity function and we end up with one over E feet per second for part C. We want to know when the particle is at rest and a particle is at rest when its velocity is equal to zero. So we take the velocity function and we set it equal to zero. We have three factors here, so we can set each factor equal to zero and solve each one so t equals zero. Each of the negative T equals zero and two minus T equals zero. When you have e to the negative T equals zero. It has no solutions because exponential functions are only positive. So the only two solutions we get R T equals zero and T equals two. So the particle is at rest at time, zero seconds and a time, two seconds. Next we want to know when is the particle moving in the positive direction? It&#x27;s moving in the positive direction when it has a positive velocity and it would be moving in the negative direction if it had a negative velocity. We just figured out when its velocity is zero. So to figure out when its velocity is positive, we can divide the number line into intervals, splitting it up at the time when it&#x27;s Ah Velocity was zero when it was at rest and so we have the interval from 0 to 2 on the interval from two to infinity. What we&#x27;re going to do is take a number from each interval and plug it into the velocity and see if we get a positive or negative. And for the first interval I chose one because we already know V of one is one over eat. So we already know that that is positive. And then for the second interval I chose three. I substituted it into the velocity and I got a negative. So what that tells us is that the particle is moving in the positive direction from time zero to time to, and we also know it&#x27;s moving in the negative direction from time to time. Infinity. All right, Next we want to find the total distance traveled from time zero to time six now because we know it&#x27;s moving in the positive direction some of the time and the negative direction. Some of the time we have to break it down into those two parts, so it&#x27;s moving in the positive direction from time, zero time to and the negative direction from time to time. Six. To find the distance, it moves in each direction. What we do is we take the difference between the position at the end of the interval and the position at the beginning of the interval. So the position at time to minus the position at time zero and we take the absolute value for each interval just in case it&#x27;s going in the negative direction and we end up with a negative difference. To find the values of F of zero f of two NF of six, you&#x27;re going to use your original function f of x or f of tea, as we called it and I decided to work with exact values. I didn&#x27;t approximate them at this point. So we substitute those numbers in and for the first distance we get the absolute value of four over a squared minus zero and for the second distance, we get the absolute value of 36 over each of the six minus 4/3 squared. We&#x27;re going to simplify those, and when we take the absolute value, what you&#x27;re going to notice about the 2nd 1 is it&#x27;s numerator is negative. And so in order to get the absolute value, you have to take the opposite of that. That&#x27;s why it looks like all I did was rearrange my numerator. Now we&#x27;re going to add these together so we get a common denominator of each of the sixth power and we combine like terms. And this is the exact value of our total distance traveled. If we want an approximate value, we can just put that into the calculator. I would get approximately 0.993 feet for part F. We want to draw the motion diagram. So now we know that the particle moves in the positive direction for a while and the negative direction for a while. So we want to show that with our motion diagram. So we start with a number line for asks the position and we start at position zero a time zero. Then we know the particle goes to the right until we get to time too. And then we know the particle goes to the left. After that part G is to find the acceleration. And remember, acceleration is the derivative of velocity. So here&#x27;s the velocity and notice that it&#x27;s a triple product. We have tea. We have times e to the negative t and we have times two minus T. And unless you really are gung ho about using the triple product rule, I would suggest that we simplify our velocity equation before we find its derivative. So what I did with I multiplied the tea and the Tu minus t together, giving me to t minus t squared. And now we can just use the regular product rule, the product of two factors. So here we have the first times, the derivative of the second plus the second time&#x27;s a derivative of the first. So I looked at that and I wanted to find a way to simplify it, and I noticed that we could factor eight. Or we could factor e to the negative t out of both terms. So I did that and that left t squared minus two t plus two minus two teak. And then we simplify that quadratic expression and we have our acceleration. Now we want to find the acceleration a time one. So we substitute one into that and we get negative one over eat and the units would be feet per second squared. Okay. For part H, we are going to use a graphing calculator and look at the graphs of the position, the velocity and the acceleration. So you grab your calculator, you go to y equals and you type those three functions in there, and then you have to think about what would make a good viewing window. So think back about some of the values you used in the problem, like we used times between zero and six, and we had some pretty small Y values things like one over E and so on. So for my window, I decided to try going from 0 to 10 on the X axis and negative 2 to 2 on the Y axis. And you could make changes to those if you don&#x27;t like them. But here&#x27;s what the graphs look like. The blue one is the original position equation. The red one is the velocity, and the black one is the acceleration. And finally, we wanted to find When is the particles speeding up? And when is the particle slowing down so the particle will be speeding up if its velocity and acceleration have the same sign, both positive or both negative. And it will be slowing down if they have different signs, one positive and one negative. So we already know about the signs of the velocity, but we don&#x27;t know yet about the signs of the acceleration. So before we confined, when the acceleration is positive or negative, we have to find when it equal zero. So let&#x27;s take our acceleration and set it equal to zero notice. We have two factors so we can set each factor equal to zero. And when we set E to the negative t equal to zero, we get no solution. And when we said T squared minus 40 plus two equals zero. Unfortunately, it&#x27;s not factory ble, so we use the quadratic formula and we end up with two plus or minus square root, too. All right, so now what we want to do is figure out the signs of the acceleration and pair that together with the signs of the velocity to find where it&#x27;s speeding up and where it&#x27;s slowing down. So I made some intervals and we have breaks at the intervals at the places where the acceleration was equal to zero, and we also have breaks in the intervals at the place where the velocity was equal to zero. Once we have those intervals established, we confined the sign of the velocity in each interval by plugging in a number from each interval, and we confined the sign of the acceleration by doing the same thing. And this is a great time to use your graphing calculator. Go to the table and you can type in different values of X. And then you confined the Y values in your Y two column and your wife. Three columns. All right, so you can see the sign results. And remember that the particle is speeding up if the signs are the same, so it would be speeding up between zero and two minus square to and between two and two plus square to and then it&#x27;s slowing down if the signs are different, so it would be slowing down between two minus square root two and two and after two plus square root, too. And that&#x27;s what we have here

Problem

Graphs of the velocity functions of two particles…

Problem 4 Hard Difficulty

Answer

a) $v=t e^{-t}(2-t)$
b) $v \approx 0.368 \mathrm{ft} / \mathrm{s}$
c) $t=0 \mathrm{s}$ and $t=2 \mathrm{s}$
d) see solution
e) see solution
f) graph unavailable
g) $a=\left(t^{2}-4 t+2\right) e^{-t}$ and $a=-0.368 \mathrm{ft} / \mathrm{s}^{2}$
h) see solution
i) see solution

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Video Transcript

Calculus: Early Transcendentals

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Precalculus Review - Intro

Functions on the Real Line - Overview

a) $v=t e^{-t}(2-t)$b) $v \approx 0.368 \mathrm{ft} / \mathrm{s}$c) $t=0 \mathrm{s}$ and $t=2 \mathrm{s}$d) see solutione) see solutionf) graph unavailableg) $a=\left(t^{2}-4 t+2\right) e^{-t}$ and $a=-0.368 \mathrm{ft} / \mathrm{s}^{2}$h) see solutioni) see solution

Calculus: Early Transcendentals

Catherine R.

Heather Z.

Michael J.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Catherine R.

Heather Z.

Michael J.

Joseph L.

a) $v=t e^{-t}(2-t)$
b) $v \approx 0.368 \mathrm{ft} / \mathrm{s}$
c) $t=0 \mathrm{s}$ and $t=2 \mathrm{s}$
d) see solution
e) see solution
f) graph unavailable
g) $a=\left(t^{2}-4 t+2\right) e^{-t}$ and $a=-0.368 \mathrm{ft} / \mathrm{s}^{2}$
h) see solution
i) see solution