# More fun with vectors

1. Feb 13, 2005

### the_d

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 2m, y = 1.5m, and has velocity Vo =(3.3m/s)i + (-7m/s)j. The accelerationis given by a = (6m/s^2) i + (5.5 m/s^2)j. What is the x component of velocity afer 1.5s?

im stuck

2. Feb 13, 2005

### vsage

A constant acceleration "a" implies the velocity is "a*t" where t is the time elapsed from 0.

3. Feb 13, 2005

### dextercioby

Put it as it should be,in vector form
$$\Delta\vec{v}=\vec{a} \Delta t$$

Now project on the Ox axis and make a simple multiplication.

Daniel.

4. Feb 13, 2005

### kdinser

First off, do you understand the problem? Can you draw it? do you know what it means that a = (6m/s^2) i + (5.5 m/s^2)j. What have you tried to do so far?

5. Feb 13, 2005

### the_d

what is the Ox axis?

6. Feb 13, 2005

### dextercioby

What do you mean?You're given that the motion takes place in the xy plane,so it's not difficult to imagine the 2 mutually perpendicular Ox & Oy axis...?

Daniel.

7. Feb 13, 2005

### the_d

i mean what is O

8. Feb 13, 2005

### dextercioby

O is the origin of the coordinate system,or if you want to,the point in which the 2 mutually perpendicular axis meet...

However,this is a useless detail for this problem...

Daniel.

9. Feb 13, 2005

### the_d

so when i get (Delta V)i subtract the initial velocity of the x component (3.5 m/s) from it and that will be the x component of velocity after 1.5 seconds.

10. Feb 13, 2005

### dextercioby

I've given you the equation already in post #3.I've explained what i meant about "Ox projection" and now i'm asking you to interpret the scalar equality
[tex] \Delta v_{x}=a_{x} \Delta t [/itex]

in a correct manner.

Daniel.

11. Feb 13, 2005

### the_d

ok i think i got it now. it would just be the acceleration of the i componet multiplied by the change in time (1.5s). thanx

12. Feb 13, 2005

### dextercioby

That will be "Delta v",to compute the final velocity (component "x") u'll have to add the initial value...

Daniel.

13. Feb 13, 2005

### the_d

curious...i did like you said, but im still not getting the correct answer. are u sure that is the correct formulae??

14. Feb 13, 2005

### the_d

okay thats where I went wrong, i dont know why i was subtracting. u were right i am supposed to add them to find the total final velocity