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Problem 3 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) = \sin (\pi t/2) $


a) $\frac{5}{2} \cos (\pi t / 2)(\text { in } \mathrm{ft} / \mathrm{s})$
b) $0 \mathrm{ft} / \mathrm{s}$
c) $t=1+2 n$
d) $(0,1),(3,5),(5,7)$ and so on
e) 6 feet
f) graph unavailable
g) $a=-\frac{\pi^{2}}{4} \sin \left(\frac{\pi t}{2}\right)$ and $a \approx-2.47 \mathrm{ft} / \mathrm{s}^{2}$
h) graph unavailable
i) $1+2 n \pi<t<2+2 n \pi$
$\mathrm{OR} :$
$3+2 n \pi<t<4+2 n \pi$


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Video Transcript

here we have a position equation and for part A, we want to find the velocity and remember that the velocity is the derivative of position. So we're going to use the chain rule and we know that the derivative of sign is co sign and then we have co sign of the inside. And then we multiply that by the derivative of the inside and then we simplify just by writing the pi over two in front of the co sign. And so then, for part B, we want to find be of one. So we're going to substitute one into that function for tea and we have the of one equals pi over two times a co sign of pi over two times one now the coast sign of pi over two times, one would be the coastline of pi over two, and the coastline of pi over two is zero. So we have pi over two time zero, which is zero, 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 is equal to zero. So we take the velocity. We set it equal to zero and we saw for teeth. So we'll divide both sides by pi. Over to now, we have co sign of pi over two t equals zero, and then we'll take the inverse coastline of both sides. So pi over two times t is the inverse coastline of zero. And that means we're looking for the angle, whose co sign is zero and they're infinitely many of them. There's pi over two. There's three pi over two. There's five Pi over two and so on now to solve for T. Let's divide both sides by pi over two and we get T equals 135 etcetera. One way we can describe that in general is if we let t equal one plus two times n where n is any integer although we want the smallest t value to be one. So we really don't want any integer. We want, uh, and to be greater than or equal to zero so we can just say in as a positive integer okay for part D. We want to know when the particle is moving in the positive direction and it will be moving in the positive direction when its velocity is zero. Now, in the previous part, we found all the times when the velocity is equal to zero. Now we want to know when the velocity is greater than zero. So what we're going to do is we're going to split the numbers into intervals, and we're going to make those splits at all the points where the, uh the lost he was equal to zero. So we're splitting it at one. We're splitting at three. We're splitting it at five, etcetera. So those are the splits for the intervals. Now we want to see what the velocity is doing on each interval. So let's pick a number in each interval and plug it into the velocity and see if we get a positive or negative. So I chose 1/2 24 and six to be the numbers I substitute into those intervals, and I got a positive, and then I got a negative, and then I got a positive, and then I got a negative. So what? That's telling me when it's positive. It tells me that the particle is moving in the positive direction, and when it's negative, it tells me that the particle is moving in the negative direction. So now we know the particle is moving in the positive direction from time 0 to 1 second from times 3 to 5 seconds, 7 to 9 seconds, and so on, going on toward infinity. Okay, Part E asks us to find the total distance traveled. Well, here's the thing. If the particle moves in the positive direction for a while and then the negative direction for a while and then positive and the negative, we've got to split this up and figure out the distance, it travels each time. Otherwise, the negative distance is going to cancel out the positive distance, so we need to find the distance. The particle travels from time 0 to 1 when it's going in the positive direction and then the distance. It travels from time 1 to 3 when it's going in the negative direction. And then, from time 3 to 5 when it's going back in the positive direction and then from time 5 to 6 when it's going back in the negative direction. And in order to do that, we're going to need a whole bunch of positions and we get those positions from our original equation F of X. So we're going to need to know the position for each and every one of these end points of the intervals 0135 and six. And a really convenient way to find those numbers without having to do all the calculating yourself is to grab your calculator. You know you're going to be graphing the function anyway when it comes to part H, so you can type the function into your calculator. Here I have why one is my original function, and I've already typed. Why to and why three. The derivative and the acceleration we'll get to those later. So once you have your original function typed into your calculator, you can go to the table and just to check your table, set real quick, go to second window and check your table set and make sure you have independent ask, because that's going to allow you to type in your own X values. So once you have that set, go into the table menu and you can type whatever X value you want. You can ignore the ones already have in here from last time and you can type X equals 1/2. I noticed the Y value is positive is well, you get an approximate value there you can type. Uh, which y values did I want? Let me go back. Okay. I didn't want 1/2. I wanted zero and one and three and five and six. Okay, so we go back to the calculator, we type zero, we get a Y value zero. We type one, we get a wide valley of one. We get three is negative. 15 is one and six is very close to zero. Looks like I'm gonna call about a zero. All right, so now that we know that to get the total distance traveled, we have to find the distance from time zero to time one. And that would be the position at time, one minus the position at time. Zero. Take the absolute value. Because when we're going in the negative direction, we're going to get a negative for this. And then to find the distance traveled from time One to time. Three. Position at time. Three minus position at time. One, etcetera, etcetera. Distance traveled from time 32 Time five. Position at time. five minus position at time three. And take the absolute value of each and every one of those before you add them all together. So we substitute our numbers in there, and then we get one plus two plus two plus one. So that gives us a total distance. Traveled of six feet for part if we're going to draw the motion diagram. So we know that this particle is going to the right to the left, to the right, to Lefferts going the positive direction, negative direction, positive direction, negative direction. So the motion diagram is going to capture that. So here we see the particle, starting at time zero at position zero. And then we see it's going to the right until we get to time one. And then we see it's going to the left until we get to time. Three. And it's going to the right until we get to time five. And then it's going to the left. Now it didn't say where to stop the motion diagram. It didn't say to stop it a time six. So if you kept going, you would just have a going back and forth back and forth back and forth part G acceleration. Remember, acceleration is the derivative of velocity. So we go back to our velocity function and we take its derivative again using the chain rule. We have a constant. And then we had the derivative of co sign was negative. Sign, negative sign of the function times the derivative of the inside. And we can simplify that and we get the acceleration is negative. Pi squared over four times the sign of pi over two times t Next, we want the acceleration at time one. So we substituted one in there. And when you're finding the sign of pi over two times one that will just be the sign of pi over two, and that is one. So we have negative pi squared over four feet per second, squared for part H. We want to use the graphing calculator and graph the position, the velocity and the acceleration. So you saw already that I had those all typed in. And you also want to think about the numbers that makes sense for this problem and choose a good viewing window. So we already saw some. Why coordinates that were numbers such as one and negative one and zero, and we saw some X coordinates, which were numbers that went up, say, zero through six and beyond. So for my window, I decided to try going from 0 to 10 on the X axis and from negative 3 to 3 on the Y axis. And you can always modify those if you don't like what you're seeing. So now we press graph, and what we see is the blue one is the original position equation. The red one is thieve a las ity equation, and the black one is the acceleration equation. Finally, we're going to find out when the particle is speeding up, and when it's slowing down, it will be speeding up when the velocity and the acceleration are either both positive or both negative, and it will be slowing down when they have different signs. One of them's positive, and one of them's negative. So we already figured out where the velocity was positive and negative, but we need to know where the acceleration is positive and negative. So first we start by finding where the acceleration is equal to zero. So we take our acceleration function and set it equal to zero divide both sides by negative pi squared over four and we're left with Sign of pi over two times T equals zero. We can take the ender sign of that. So what we're looking for would be the angles that have a sign Value of zero and the angles that have assign value of zero would be zero pi to pie three pie for pie all the multiples of pi. Now, to get t by itself, we divide both sides of that by pi over two and we end up with t equals 02468 etcetera All the evens. So we could say that t is to in and in this case Ah, the smallest end can be a zero. So we'll say in is the whole number in a 0123 etcetera. Okay, so now what we want to do is break up our number line into intervals. We have to break it at one and at three. And at five. And at seven. And at nine. Because those were the places where the velocity changed sign. But we also have to break it up at two and four and at six. And so on because those are the places where the acceleration is changing. Sign. So what we want to do then, is analyzed the sign of the velocity on the sign of the acceleration on each of those intervals and determine if we have the same sign for speeding up or opposite signs for slowing down. And this is another great place to use your calculator If you go back to the table and my Y two column will be my velocities and my wife three column will be my accelerations, and what I could do is just take numbers from each interval and type them in, and I'll be able to find out whether the velocity is positive or negative and whether the acceleration is positive or negative. So I did that, and I have all the signs here. So based on that, we know that it's slowing down when the signs are different. So it's slowing down from time zero to time, one from time to time, three from time for two time five and so on and so on and so on, and it's speeding up when the signs of the same so it's speeding up from time. One to time. Two from time. 32 Time for from time. 52 times six etcetera. So if you're looking for ah algebraic way to express that, we can say that it's slowing down from time to in to time to in plus one where in is a whole number, not an integer, and it's speeding up from time to n plus one to time two in plus two.

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