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