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A particle moves according to a law of motion $ s…

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Problem 2 Hard Difficulty

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) = \frac {9t}{t^2 + 9} $

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Video Transcript

here we have a position equation and for part a of the problem. We want to find the velocity. Remember that velocity is the derivative of position. And because our position equation is a quotient, we're going to find the derivative using the quotient rule. So here we have the bottom times, the derivative of the top, minus the top times the derivative of the bottom over the bottom squared. Next, we need to simplify that. So what I've done is distribute the nine and multiply the 90 and the two t And then I combined the like terms. So that's our velocity for part B. We want to find the velocity at time one. So we're going to substitute a one in for tea and then simplify and we end up with 18 25th and the units would be feet per second for part C. We want to know when the particle is at rest and a particle is at rest if its velocity ease is equal to zero. So we take our velocity, we set it equal to zero, and we saw for time. Now, when you have a fraction that equal zero, you want the numerator to equal zero. And when you look at this denominator, you'll notice that there's no real number that makes it equal to zero anyway. So we set the numerator equal to zero, and we saw for tea we can add 90 square to both sides. Now we have 90 squared equals 81. We can divide both sides by nine. Now we have t squared equals nine and weaken square root. And typically we would write plus or minus three. But because for this problem we're starting at time zero and only using positive times, we're not going to include negative three. Okay, so now we know the particle is at rest at time. Three seconds for part D. We want to know when is the particle moving in the positive direction and it will be moving in the positive direction if the velocity is zero. So what we do is we take the information from the previous part. We found out that the velocity was zero when the time was three. And what we do then is we make test intervals before three and after three to figure out what the velocity is doing. So before three, we have the interval from 0 to 3 and after three, we have the interval from 3 to 0. So what we want to do is pick a number in each interval and test it into the velocity to see if we get a positive or negative. So I picked one, and I picked for and remember, back in the previous part of the problem, I think it was part B. We found that V of one was positive 18 25th. So we know that via one is positive, and then if you substitute for into the velocity, you get a negative. So what this is telling us is that the particle is moving in the positive direction from time 0 to 3, and it's moving in the negative direction from time. Three to Infinity for party. We want to find the total distance traveled by the particle from time zero to time. Six. So here's what we just learned. We learned that it's traveling in the positive direction from time zero to time three and then it's traveling in the negative direction from time 32 times six. So what we don't want to have happen is we don't want a negative to cancel out the positive distance. And so we're going to take the absolute value of each distance. We're gonna break it down into the distance from time zero to time three and the distance from time 32 times six and take the absolute value each time. So to get the distance from time zero to time. Three. What we do is we find the position at time three minus the position at time zero and then we take the absolute value and to get the distance from time 32 times six. We take the position at times six, minus the position a time three and then take the absolute value. Now those position values come from the original function F of X. So if you need to find F zero F of 3.5 of six, you can substitute those into the your F of X function, and you can use the table feature of your graphing calculator to do that. It's really helpful, so those numbers go in. So we have the absolute value of 1.5 minus zero, plus the absolute valuable in 00.2 minus 1.5, and that gives us the absolute value of 1.5 plus the absolute value of negative 0.3. And so we have 1.5 plus 0.3. So the distance travelled is 1.8 feet. Now we're going to make a motion diagram and because we know it went in the positive direction for a while and in the negative direction for a while, we should see that in our motion diagram. So what we see is a particle starting at time zero position zero. It moves to the right and it stops moving to the right a time three. And then it moves to the left for part G. We want to find the acceleration. And remember, acceleration is the derivative of velocity. So we had our velocity equation. Now we take the derivative of it using the quotient rule, and we get our acceleration equation. So here we have the bottom times, the derivative of the top minus the top times, the derivative of the bottom, using the chain rule to do that over the bottom squared, that's going to need to be simplified if we're going to be able to use it for anything So I distributed and multiplied these together. Actually, no, I did not distribute. What I did was I factored out a t squared plus nine from here and a T squared plus nine from here. And I canceled it with one of the t squared plus nine on the bottom. So that left me with what we see in the next step. And then I distributed. So I multiplied the 18 t by t squared plus nine and I multiplied the fourty by 19 minus 90 squared. And once we have that, we can combine the like terms. And then I decided after that that I would factor out the common factor of the numerator which was 18 t because I'm anticipating that eventually we're going to get to the point where we have to set this equal to zero. And I thought it would be easier if it was factored. Okay, The next thing we want to do is find the acceleration a time one. So we substitute one into that acceleration function and we can simplify it and we get negative 1 17/2 50 the units would be feet per second squared. Now that we have the position, the velocity and the acceleration. We can use a graphing calculator and look at all three graphs. So we go to the calculator, we go to the y equals menu, we type them all in, and now we need to set a good viewing window that's going to allow us to see all of these. So think back over some of the numbers we got from the previous parts of the problem. Like we found out that the positions were numbers like zero on 1.5 and 1.2. We found out that times were numbers like zero and three and six. So based on that, I went to a window and I decided to try letting my ex values go from the negative 1 to 10 and letting my Y values go from negative 2 to 3. And you can also always fiddle with these numbers and make changes until you like what you see. So what we have here, the blue one is the graph of the position equation. The red one is the graph of the glossy equation on the black one is the graph of the acceleration equation. Okay, After that The final part is to find out when the particle is speeding up and when it's slowing down and a particle will be speeding up if its velocity and acceleration are both positive or both negative and it will be slowing down if its velocity and acceleration have opposite signs, one with a positive one with a negative. So we need to figure out when our acceleration is zero so that we can then figure out when it's positive and when it's negative. So we take our acceleration and set it equal to zero, and you'll notice here that I just have the numerator because if a fraction is equal to zero, it's just the numerator that's going to equal zero. And so then we can set each factor equal to zero, and we can solve each of those and we end up with T equals zero or T equals three square root three. And we're not including the negative time, of course, because we're only talking about times starting at zero and going forward. Okay, so what we're going to do once we know the times that the acceleration is equal to zero is break the number line down and we're going to find the acceleration between time zero and time three Square root three and we'll find the acceleration between time three square three on infinity. So what I did was I plugged in a number into the acceleration. I plugged in a number that was in the first interval. We had already plugged in a one, so that's convenient and it turned out negative. And then I took 10. I never That was in the second interval. Put it into the acceleration and I got a positive. So now let's let's bring that information together with the information from the velocity. So I'm breaking down the intervals and I've broken down the intervals at three square Root three because that is where we see the change in acceleration. But I've also broken down the intervals at three because that's where we saw the change in velocity. So what we saw with velocity was it was positive between zero and three, and it was negative after that. What we saw with acceleration was that it was negative before three square or three, and it was positive after that. So based on that, when the signs are both the same negative and negative. We know the particle is speeding up, so it's speeding up from times 3 to 3 square root three. And when the signs are different, a positive and a negative, we know that the particle is slowing down, so it's slowing down between times zero and three and times three, square three and infinity.

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