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A particle moves according to a law of motion $s=f(t)$ $t \geqslant 0,$ where $t$ is measured in seconds and $s$ in feet.(a) Find the velocity at time $t$.(b) What is the velocity after 1 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 $0 \leqslant t \leqslant 6$(i) When is the particle speeding up? When is it slowing down?$f(t)=t^{3}-8 t^{2}+24 t$

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00:14

Felicia Sanders

06:44

Heather Zimmers

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Monicah W.

December 11, 2021

Given that the curve y=9-12x+x^ ^ 3 has two turning points, determine i. Derivative of y with respect to x ii. Value(s) of x at which the curve is stationery iii. Coordinate at the lowest turning point iv. Coordinate at the highest turning point

Catherine A.

October 27, 2020

That was not easy, glad this was able to help

David Base G.

October 23, 2020

Doug F.

September 23, 2020

This is just easy Brad the direction of motion is the direction the velocity vector points in. To take your example of a projectile fired straight up, because only it's height, z, is changing the velocity vector will look like: ?v=(0,0,dzdt).

Brad S.

What is motion direction?

Eric V.

In calculus, the acceleration is found as being the derivative of velocity. In other words an object whose velocity is v has an acceleration of which is read as "dv - dt", and is the rate of change of the velocity v, with respect to time. I hope this make

Sarah H.

What is acceleration?

Erica G.

Hey Julia I think quadratic equations are actually used in everyday life, as when calculating areas, determining a product's profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the

Julia M.

What is the use of a quadratic equation?

Oregon State University

Harvey Mudd College

University of Nottingham

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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A particle moves according…

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31:04

04:05

06:42

particle moves according t…

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26:13

$1-4$ A particle moves acc…

09:02

08:32

Alright. So here we're going to look at a object moving uh with a motion graph or position graph um of F A. T. And the the position with time on the straight line is T cubed minus 80 squared plus 2014 where T. Is in seconds and ss and feet. And we're only interested for t greater than or equal to zero. And so what we're gonna do is based on this motion equation, we are going to solve the answer a lot of questions. So let's go to it. Okay, first of all, we want to find V. F. T. And V A T. Is a derivative of our position graph. So we're going to go ahead and use power rule, we'll get three T squared minus 16 T Plus 24 by power roll. And our units will be feet per second. So in feet per second. Alright, so in part B We want the velocity at 1/2. So that would be just what happens when you plug in one for T. So we'll go ahead and plug in one And that will give us 3 -16 plus 24 or 11 feet per second. So that is I'll just box our answer so far. All right, let's take a look at part see part C. Is we want to find out when our object comes to rest. So we need V. F. T. Which is three T squared minus 16 T Plus 24 to equal zero. And then the goal is to solve for T. So not necessarily easy to factor when you have that three in front. Let's go ahead and use quadratic formula. So quadratic formula says then t will be equal to -7. Remember we've got a the and see. So T. Is minus B. So 16 plus or minus square root of B squared so minus 16, quantity squared minus four times a time. See And that's all over two. A. so over six. The trouble is is that um when we plug in um to our calculator we end up, I just kinda leave room here we get 16 plus or minus. Oh no a minus 32 square reading a negative gives us an imaginary answer. Therefore there are no zeros that are real. Therefore VF T never goes to zero. Therefore we can say um the object mm hmm never uh is at rest. Okay, so VF T never goes to zero. Okay, so that's good to know. Or let's keep going up top. Next thing we need to know is when is the particle movie in the positive direction? Well, positive direction ah corresponds to velocity being positive. Well, I'll just say being positive, not just positive. Alright, since our velocity we know the object is never at rest, we can just plug in any time. And if we get a positive value is always that way. So let's plug in T. A zero and it's positive. So since Sense v. of zero. Yes uh greater than zero and we never stop. That means we're always going in the same direction. So therefore we are always uh the object is always going in the positive direction. Okay. Good to know. Alright let's keep going. We still have some more to go. Alright so part E. We need to find the total distance. And since we're always going in the same direction, the total distance is really just the change in our position, so position at and we're doing it over the first um six seconds. So the difference between where we're at at six seconds and where we're at at zero seconds. So at six seconds we plug in six store equations. So we're going to get six cubed -8 times six squared Plus 24 times six. And then all the terms have T. Units, so S. Of zero is zero And when we plugged that in our calculator we end up with 72 ft. So we traveled 72 ft in those first six seconds. So this is um For six seconds, first six seconds. Okay, excellent. Alright we can keep going, we have lots of parts. Okay. Part you want to kind of sketch what's happening and because we're always going in the same direction, we do start at zero, This is zero and we do know that at Where we reach 72 ft at T equals six seconds at T. equals zero seconds were at zero ft. Um Okay so we basically are just always moving in the positive direction and so on. So that's pretty easy graph there. Alright. Still three more to go, I might have to clear the screen. Let's see if I can write it. Super squishy. Okay let's look at G. She is finding the acceleration um and then plugging it in one second so that's the next derivative. So if we do power roll on V We'll end up with 60 -16. So if we want a of one, That'll be six times 1 -16. So we get -10 but our units are feet per second squared, so minus 10 feet per second squared. Um Okay, so I'll probably might need to race part of it. Let's see if I can at least do uh the best I can. I'm going to make a little spot for the remainder and this is where um Well I think I'm actually going to have to erase so I will erase what I have a race to make space because we still have three more parts. So I will erase after just a sec here. Okay, so I cleared the screen and then rewrote for us the position velocity and acceleration equations part H asks us to graph rough sketch of our position velocity and acceleration graphs. So I did it rough just so you can see the uh with time roughly the kind of shapes of each of these graphs. And then we've got still part I and part J. Part I says basically we have to find out um well I and J will kind of do together um I is basically times when object is speeding up and jay is times when object is slowing down. Okay, so um the rule for objects speeding up is when velocity and acceleration have the same sign and slowing down is when velocity and acceleration have opposite signs. Alright, so we already learned from previously that we have always a positive velocity. Um and you can see from the little graph in blue that velocity never hit zero or goes negative, it's uh stays positive. So since V is greater than zero than for speeding up, We need acceleration to also be greater than zero and on the opposite case for slowing down since V is greater than zero then acceleration has to be negative. So if you look at our graph of acceleration, you can see we're negative up until this point here and then it switches to positive. So we just really need to know when a of T. Goes to zero. So since a of T Equal 60 -16, we can set that equal to zero. You can add 16 to both sides and then divide by six. So that's 16/6 seconds. Or if I divide by 28 3rd 2nd. So basically um when we're below eight third seconds then acceleration is negative. So we um if we're less than eight third seconds then we have a we're slowing down and I think we're only counting, starting at zero. So this is slowing down and after that the acceleration is positive. So after that we are speeding up and I suppose I could probably Um just do it without the equal sign. So T greater than eight 3rd 2nd. We are speeding up and I'll go ahead and I think uh I think this is good. So we have accomplished all the parts for our motion. Okay, have a wonderful day and I'll see you next time.

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