# Velocity Maximum

1. Sep 17, 2009

### tjbateh

1. The problem statement, all variables and given/known data

A particle leaves the origin with an initial velocity v = (7.44i) m/s and a constant acceleration a= ( - 2.60i - 2.70j) m/s2. When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector?

2. Relevant equations

3. The attempt at a solution

I have found that the Magnitude of Acceleration is 3.75.

What other information do I need to find to solve this problem?

2. Sep 17, 2009

Hello

Was this all that was given? You will need to find 'what' that maximum x-coordinate is.

Can we find that from the given information?

3. Sep 17, 2009

### tjbateh

I believe we can find that information, but I do not know how.

4. Sep 17, 2009

I belive we are short some info here. This is some variation of projectile motion yes?

Is there a height that this object is falling that was given?

EDIT: oops! Note the directions of acceleration and velocity in the i direction.

What can you say about the i-component of velocity when it has reached its max x-coordinate?

5. Sep 17, 2009

### tjbateh

when it reaches its max X-coordinate, y=0??

6. Sep 17, 2009

I think you need to reread what I wrote

7. Sep 17, 2009

### tjbateh

it is 7.44? I'm confused.

8. Sep 17, 2009

### tjbateh

ok for part A I got -7.74j m/s^2.

for T (time) i got 2.865.....now how do I get the new position vector?

9. Sep 17, 2009

Hey there

The time 't' that I got is a little different.

$$(v_f)_x=0=(v_o)_x+a_xt$$

$$\Rightarrow t=\frac{(-v_o)_x}{a_x}=\frac{-7.44}{-2.60}=2.862\text{ s.}$$

Using this 't' in the y-direction to find vy we have

$$(v_f)_y=(v_o)_y+a_yt$$

$$\Rightarrow (v_f)_y=0+(-2.70)(2.862)=-7.23\, \mathbf{j}\frac ms$$

For the position vector, you can easily find the maximum x-coordinate with the given information and I will also assume that you can find its y-coordinate at time 't' = 2.862 s, right?

So if you have both of its coordinates, surely you can come up with its position vector:

$$\mathbf{r}=<(x_f-x_o),\, (y_f-y_o)>$$

10. Sep 17, 2009

### tjbateh

Great help, but I really do not know how to find the maximum x-coordinate. I have yet to understand that concept.

11. Sep 17, 2009

Kinematics homey! You know your kinematic equations right? (or at least have a text where you can look them up ).

You have acceleration, initial velocity in x, final velocity in x (at its max x-coordinate). Egads man! You even have 't' !

You can use almost any equation you like

Last edited by a moderator: May 4, 2017
12. Sep 17, 2009

### tjbateh

Ok I have all of that, but which one is the final velocity of x?

13. Sep 17, 2009

Really? How the heck did we get all the way to this point without your understanding that?

Is it just me? Or did I not already have this conversation with you:

If you didn't understand this, how did you get t = 2.8 and V = -7.74 j ?

14. Sep 17, 2009

### tjbateh

I set the x component of velocity equal to zero to solve for T. then used that T and plugged it into the equation of the y component of velocity to get -7.74j...I got all that but for some reason I am lost.

15. Sep 17, 2009

16. Sep 17, 2009

### tjbateh

Ok, that gives me T. How is that my answer? So the final of y is equal to -7.74...and the final of x? What is that?

17. Sep 17, 2009

Come on buddy. Work with me. Why would the position vector be equal to the velocity? Does that even make sense?

I told you explicitly, step by step, exactly how to solve this problem:

Find the final y-coordinate for me. SHOW me all of the steps.

You have $(v_f)_y,(v_o)_y,a_y,y_o \text{ and }t$

From kinematics:

$(v_f)_y^2=(v_o)_y^2+2a_y(y_f-y_o)$

Do it.

18. Sep 17, 2009

### tjbateh

Ok so for Vy I got -.2874....So I get -7.73j for the y-component of the position vector. I am sorry, but I still seem to be having trouble getting the max X component. I know once I find that I can find the x component of the position vector.

And from the right side of that equation at the very beginning, I got 41.74. Is there a variable in that equation I am supposed to solve here?

I'm trying here, I appreciate ya sticking with me on this!

19. Sep 17, 2009

Again, I need you to SHOW all of steps for this. This does not make sense. We ALREADY got Vy = -7.73 j m/s ; what is this "-.2874" number?

Think about it; does a number less than 1 make sense? The acceleration in the y-direction was given as -2.70 m/s2 AND we know that at the point in question, the particle has been traveling for t = 2.862 s.

Therefore to find Vy at t = 2.862 s we have

$(v_f)_y=(v_o)_y+a_yt$

$\Rightarrow (v_f)_y=0+(-2.70)(2.862)$

$\Rightarrow (v_f)_y=7.73$ m/s (3) But as I said, we already did this in post#9 !

Now using the equation I gave you in post#17 to find the y-coordinate :

$$(v_f)_y^2=(v_o)_y^2+2a_y(y_f-y_o)$$

$$\Rightarrow y_f=\frac{(v_f)_y^2-(v_o)_y^2}{2a_y}+y_o$$

$$\Rightarrow y_f = \frac{7.73^2-0^2}{2(-2.70)}+0$$

$$\Righarrow y_f =-11.07\, \mathbf{j}\text{ m}$$

Now let's use the exact same equation to find the x-coordinate:

$$(v_f)_x^2=(v_o)_x^2+2a_x(x_f-x_o)$$

$$\Rightarrow x_f=\frac{(v_f)_x^2-(v_o)_x^2}{2a_x}+x_o$$

$$\Rightarrow x_f=\frac{0^2-(7.7)^2}{2(-2.60)}+0$$

$$\Righarrow x_f =10.59\,\mathbf{i}\text{ m}$$

Therefore, since you started at the origin, the position vector is simply the vector that points at your new final coordinates:

$$\mathbf{r}={10.6\mathbf{i}-11.1\mathbf{j}}\text{ m}$$

I really hope that you do not just copy this answer down and call it a day. I suggest going back over this entire thread tomorrow with pencil and paper in hand and actually do all of the steps.

That is the only way you will master these kinds of problems

If you have any questions about any of the values I plugged into the equations, just ask.

20. Sep 18, 2009

### tjbateh

Thank you so much for going through all those steps. What was confusing me was the variables in these equations. It was difficult for me to point out which one was X final, and which one was Y final. I must have not been looking at it right. It makes a lot more sense now, and I worked it out on paper. I will need a lot more practice though before I get the hang of it! thanks again!