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Problem 1 Hard Difficulty

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?
$f(t)=t^{3}-8 t^{2}+24 t$

Answer

a) $v(t)=3 t^{2}-16 t+24$
b) 11 $\mathrm{ft} / \mathrm{s}$
c) The particle is never at rest.
d) The particle ALWAYS moves in the positive direction.
e) 72 feet
f) Go to http://m.wolframalpha.com/input/?i=ln%28x%29&x=0&y=0 and type in ln(x) (only look at the values of x greater then 0 or in other words choose the real option) and then also graph 1/x.
g) $a(t)=6 t-16$
After 1 second: $-10 \mathrm{ft} / \mathrm{s}^{2}$
h) graph unavailable
i) Speeding up: $t > \frac{8}{3}$
Slowing down: $0 \leq t < \frac{8}{3}$

More Answers

Discussion

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JM

Julia M.

September 23, 2020

What is the use of a quadratic equation?

EG

Erica G.

September 23, 2020

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

SH

Sarah H.

September 23, 2020

What is acceleration?

EV

Eric V.

September 23, 2020

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

BS

Brad S.

September 23, 2020

What is motion direction?

DF

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).

DG

David Base G.

October 23, 2020

That was not easy, glad this was able to help

CA

Catherine A.

October 27, 2020

That was not easy, glad this was able to help

Video Transcript

here we have the equation for the position of the particle and for party. What we want to do is find the velocity and remember that velocity is the derivative of position. So we take the derivative of F of X, and we get three t squared minus 16 T plus 24. Next, we want to find the velocity at one second. So we're finding V of one. So we substitute one into the velocity equation that we just found in part a and then we simplify and we have 11 and the units would be feet per second. Next, we want to know when the particle is at rest and a particle will be at rest when its velocity is equal to zero. So we want to take our velocity and set it equal to zero. Then we saw for teeth. So this particular quadratic equation is not factory ble. So we end up using the quadratic formula and the solutions air non riel. So that means there are no times when the particle is at rest. Next, we want to know when is the particle moving in the positive direction and it will be moving in the positive direction whenever the velocity is positive. Now we found in part B that the velocity is positive at time. One. It was positive 11 and we found that there were no times when the velocity equals zero. So that means it's either always positive or always negative. And since we know that it's positive a time one we know the velocity is always positive. So the particles always moving in the positive direction. Now we're going to find the total distance traveled from time one to time. Six. So because the particle is moving in the same direction the whole time, all we have to do is find the position of the particle a time six and the position of the particle a time zero and subtract. Hm. And we get those positions from our original position equation. So a substitute six into the original equation and we get 72 and we substitute zero into the original equation and we get zero. So we're subtracting zero from 72 that gives us 72 feet as the total distance. Next, we're drawing emotion diagram, so we draw a horizontal axis for the position s we start at time zero, which had a position of zero. So that's our starting point, and we end at times six, which had a position of 72 were showing the arrows showing that we're going in the positive direction and it didn't turn around. So it's a pretty simple motion diagram. Next we're finding the acceleration and acceleration is the derivative of position. So we take the derivative. Excuse me. Acceleration is the derivative of velocity. Let's change that. There we go. That's better. Acceleration is the derivative of velocity. So we take the velocity equation, find its derivative and we get 60 minus 16 and that's acceleration. And then we went to find the acceleration at time one. So we substitute one into that and we get negative 10 and the units would be feet per second squared. The next thing we're going to do now that we have the position equation, the velocity equation and the acceleration equation, is we're going to look at the graphs of all three. So we grab a calculator, we go into y equals and we type in the functions. So the 1st 1 was the position function, the cubic function, the 2nd 1 is the velocity equation, and the 3rd 1 is the acceleration. Now you're going to need to play around with your window dimensions to get a good view. And you can think about the numbers that you've already seen in the problem. For example, you're looking at times from 0 to 6 you've seen a point that had a height of 72 another point that had a height of zero. So take all that into account and what I tried was my ex values going from negative 1 to 10 and my wife values going from negative 50 to 100 and I did play around and adjust those till I found something I wanted. So here we see that the blue one is the position function and the red one is thieve a las ity function, and the black one is the acceleration function. Okay, the last thing we want to dio is, look at when is the particle speeding up and when is it slowing down? So when the velocity and the acceleration are both positive or both negative, it's speeding up. And when the velocity and acceleration are have different signs one positive one negative it's slowing down. So from Part D, we learned that the velocity is always positive. So now we need to find the sign of the acceleration. So what we do is we take the acceleration and we set it equal to zero. And once we know where it's equal to zero, we can figure out where it's positive and where it's negative. So when you set the acceleration equal to zero and solve for T, you get 8/3. After that, you make intervals and you have an interval from 0 to 8/3. And then you have an interval from 8/3 to infinity and you take a number from the interval. You plug it into the acceleration and you find out if you get a positive or negative. Remember earlier in the problem, I found that the acceleration at one was negative. 10. So I took advantage of that idea, and I know that the acceleration is negative on the first interval, and then I took a number greater than 8/3. Maybe it could be four or 10 or 20 or something substituted it into the acceleration and found that I got a positive. All right, so Let's put that together with our velocity information, and we know the velocity is always positive. So if the velocity is positive and the accelerations positive, the particle is speeding up. So that's happening when T is greater than 8/3 and then when the velocity is positive and the acceleration is negative, the particle is slowing down, and that's happening when time is between zero and 8/3 seconds.

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