# Homework Help: Homework, Particle Question

1. Nov 16, 2006

### XPX1

A particle passes through the origin with an initial speed of 5.6 m/s in the positive y direction. The particle accelerates in the negative x direction at 5.2 m/s2.
(a) What are its x and y positions after 4.0 s?
m (x position)
m (y position)
(b) What are vx and vy at this time?
m/s (vx)
m/s (vy)

This question seems pretty hard to me, I have tried about 4 times to get the answer to no avail, heres what I have done.

To try and find out where its x and y positions were after 4.0 seconds, I plugged it into the Position Versus Time Equation, xf=xi+vixt+.5axt^2

Since I have the givens

time = 4.0
Vi=5.6
a=-5.2^2

I plug it into the equation.

xf=xi+vixt+.5axt^2

xf=0+5.6(4)+.5(-5.2)^2(4.0)^2

xf=238.72

however, this answer for part A. was not correct.

Can anyone tell me what I am doing wrong here?

2. Nov 16, 2006

### conquertheworld5

You have to treat each direction separately.
you will have two equations
eq 1: xf=xi+vixt+.5axt^2
eq 2: yf=yi+viyt+.5ayt^2
think about what the question says and in which direction each velocity and acceleration is in.

3. Nov 16, 2006

### XPX1

Aaahh.. I see, so if it is going in the x direction at -5.2m/s^2 then after 4 seconds, it would be xf=0+0+.5(-5.2)^2(4)^2?

and if it is going in the y direction

yf=0+5.6(4)+.5(?)(4)^2

So, how do I figure out the y if I am not given the acceleration?

4. Nov 16, 2006

### conquertheworld5

If you're not given an acceleration, I think you can assume that there is none and the particle is moving at a constant velocity in the y direction.

5. Nov 16, 2006

### XPX1

I got the final position of the y correct, but for x I am still lost.

if X is accelerating at -5.2m/s^2, where will it be in 4 seconds?

I plugged it into the position versus time equation

xf=0+0+.5(-5.2)^2(4)^2
xf=216.32

The above answer was false, I am completely lost on this, thanks for your help on solving the y position though conquer!

6. Nov 16, 2006

### conquertheworld5

xf=xi +vixt +.5a(t^2)
"xf=0+0+.5(-5.2)^2(4)^2"
you have something I don't...

7. Nov 16, 2006

### XPX1

what am I missing? I still dont see what I have that you dont

8. Nov 16, 2006

### conquertheworld5

heh, take a look at the acceleration...after everythign cancels, you have
xf=.5(a^2)(t^2)
I have
xf=.5(a)(t^2)

9. Nov 16, 2006

### XPX1

But.. its 5.2 meters a second squared, so dosnt that make it 5.2^2? or do I take the square root of 5.2 and put it into a?

xf=.5(-5.2^2)(4^2)

or...

xf=.5(2.28)(4^2)?

10. Nov 16, 2006

### conquertheworld5

okay, here's the units:
acceleration = m/s^2
t^2 = s^2
so
a x t^2 = m/s^2 x s^2 = m

so it's just xf = .5(-5.2)(4^2), this will give you your answer in meters.

a is already in m/s^2, if you square it again, your units would be m^2/s^4

11. Nov 16, 2006

### XPX1

Ah thanks, that was correct! I still dont quite understand why you dont leave it m/s^2 though

12. Nov 16, 2006

### conquertheworld5

hmm... I'm not sure what you mean by this.
You ARE leaving the acceleration in m/s^2 because by definition, acceleration is in m/s^2. Doing anything to that acceleration would change the formula.
You don't want the final answer in m/s^2 because you have a distance, x, which should be in meters, on the other side of the equation.

maybe there's someone else in the forums who can explain this more clearly... anybody care to give it a try?

13. Nov 16, 2006

### XPX1

I think I understand it now, thanks for all your help!