# Vector components of velocity

1. Oct 19, 2009

### waldvocm

1. The problem statement, all variables and given/known data
A particle starts at t=0 at the origin of a coordinate system with a velocity of 1.32m/s in the negative y direction. There is acceleration on the particle of 2.47m/s^2 in a direction 35 degrees above the x axis.

a) at what time does the y component of the particle's position equal 3.38
b)At that time, what is the x component of position
c)at that time, what is the x component of velocity
d)at that time, what is the y component of velocity

For a)........

I used Rf=Ri+Vit+1/2at^2

3.38=0+-1.32(t)+1/2(2.47)t^2
My final answer is t=2.27 Is this correct

For b)........

I used Xf=Xi+vxiT+1/2at^2

Xf=0+-1.32(2.27)+1/2(2.47)(2.27)^2
My final answer is 3.37 Is this correct

Last edited by a moderator: Oct 19, 2009
2. Oct 19, 2009

### waldvocm

ok, for c)....
Vx=4.29cos(35)=3.51

For d)
Vy=4.29sin(35)=2.46

Is this the right way to complete the steps for this problem

3. Oct 19, 2009

### waldvocm

or.....

Do I take the derivative of the position equation to find velocity?

4. Oct 19, 2009

### Redbelly98

Staff Emeritus
Oops. I have accidentally edited part of your post #1. My apologies!

No.
Since this equation is for y, what is the y-component of the acceleration?

No.
This equation is for x. So, what are the:
• x-component of the initial velocity, and
• x-component of the acceleration?

5. Oct 19, 2009

### Redbelly98

Staff Emeritus
I don't understand this; where does the 4.29 come from?

You could do that, or, equivalently, just use the kinematic equation for velocity vs. time.

6. Oct 20, 2009

### waldvocm

Here are my components.....
Vyi=-1.32 Vxi=0

ax=2.47(cos(35))=2.02 ay=2.47(sin(35))=1.42

a) 3.38=0+-1.32(t)+.5(1.42)t^2= .71t^2-1.32t-3.38 t=3.30

b) Xf=0+0(3.30)+.5(2.02)(3.30)^2=11.0

c)v=0+2.02(3.30) =6.67

d)v=-1.32+1.42(3.30)=3.37

Does that look better?

7. Oct 21, 2009

### Redbelly98

Staff Emeritus
Looks good!