more fun with vectors

[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

[vsage]

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

[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.

[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?

[the_d]

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.


what is the Ox axis?

[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.

[the_d]

i mean what is O

[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.

[the_d]

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.

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.

[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.

[the_d]

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.

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

[dextercioby]

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

Daniel.

[the_d]

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

[the_d]

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

Daniel.

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