Finding Final Velocity with Constant Acceleration

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The discussion focuses on calculating the final velocity of a particle moving in the xy plane with constant acceleration. The initial velocity and acceleration are provided, and participants clarify the use of vector equations to find the x component of velocity after 1.5 seconds. Key points include the importance of correctly interpreting the scalar equality for velocity change and the necessity of adding the initial velocity to the change in velocity to find the final velocity. Misunderstandings about the coordinate system and calculations are addressed, leading to a resolution of the problem. Ultimately, the correct approach involves using the acceleration and time to find the change in velocity and then adding it to the initial velocity.
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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
 
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A constant acceleration "a" implies the velocity is "a*t" where t is the time elapsed from 0.
 
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.
 
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?
 
dextercioby said:
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?
 
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.
 
i mean what is O
 
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.
 
dextercioby said:
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.
 
  • #10
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
\Delta v_{x}=a_{x} \Delta t [/itex]<br /> <br /> in a correct manner.<br /> <br /> Daniel.
 
  • #11
dextercioby said:
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
\Delta v_{x}=a_{x} \Delta t [/itex]<br /> <br /> in a correct manner.<br /> <br /> Daniel.
<br /> <br /> 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
That will be "Delta v",to compute the final velocity (component "x") u'll have to add the initial value...

Daniel.
 
  • #13
curious...i did like you said, but I am still not getting the correct answer. are u sure that is the correct formulae??
 
  • #14
dextercioby said:
That will be "Delta v",to compute the final velocity (component "x") u'll have to add the initial value...

Daniel.

okay that's where I went wrong, i don't know why i was subtracting. u were right i am supposed to add them to find the total final velocity
 
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