# Homework Help: Uniform object acceleration

1. Feb 26, 2008

### chocolatelover

1. The problem statement, all variables and given/known data
An object moving with uniform acceleration has a velocity of 16.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 1.60 s later is -5.00 cm, what is its acceleration?

2. Relevant equations
a=vf-vi/change in time

3. The attempt at a solution

a=3--5/1.6 s

Could someone please tell me if this looks correct?

Thank you very much

2. Feb 26, 2008

### XxBollWeevilx

It looks like you are using your coordinates as the velocities...this would not be correct. Think in terms of displacement. Do you know any formulas that can be used with uniform acceleration?

3. Feb 26, 2008

### belliott4488

It always helps to include the units of all the numbers in your equations. You're subtracting two numbers that have units cm and dividing them by something in seconds, so your result will be in cm/sec. That's a speed, however, not an acceleration.

As XxBollWeevilx pointed out, you need to use a different equation - one that combines just the quantities you're given and the one you're trying to solve for.

4. Feb 26, 2008

### chocolatelover

Thank you very much

Does this look righ?

displacement=xf-xi

so, 3cm--5cm=displacement
=8cm

The displacement is just the distance, right?

The acceleration is:

vf-vi/change in t

I don't understand how the displacement helps. I still don't know what vi is, right?

Thank you

5. Feb 26, 2008

### cse63146

distance is how much you travelled, displacement is the difference between the initial position of a reference point and any later position.

6. Feb 26, 2008

### XxBollWeevilx

You're going from 3 cm to -5 cm, so your displacement will be $$x_f-x_i$$ = -5 cm - 3 cm. You DO know what vi is, from what I can see, but you don't know vf. But you don't need vf. Are there any equations that do not require vf? Think kinematics.

7. Feb 26, 2008

### chocolatelover

Thank you very much

So, if the displacement=xf-xi or 3cm--5cm=8cm in this case, then I can find the velocity, right?

velocity=xf-xi/change in t

8cm/1.6-0s=5m/s

In order to find the acceleration, I need to take vf-vi/change in t and it needs to be in m/s^2, right?

I know that the acceleration in uniform or constant. I know that vf=vi+at, but I don't know what vi or a are. Could you show me what I need to do to find the acceleration?

Thank you

8. Feb 26, 2008

### cse63146

its -5 cm - 3 cm not 3cm--5cm

displacement can be negative, distance cant

9. Feb 26, 2008

### XxBollWeevilx

No, don't worry about finding velocity. Taking xf-xi/change in t will only give you the average velocity during the motion, not the initial or final velocity. Look for a formula that doesn't require vf and everything but a is known.

10. Feb 26, 2008

### chocolatelover

Thank you very much

Could I use this equation?

xf=xi+vit+1/2at^2?

Thank you

11. Feb 26, 2008

### XxBollWeevilx

That would be a very good equation to use! :)

Nice job.

12. Feb 26, 2008

### chocolatelover

Thank you very much

Does this look correct?

-5=16+16(1.6)+1/2a1.6^2
a=-36.41

Thank you

13. Feb 26, 2008

### XxBollWeevilx

I didn't check the arithmetic, but that looks pretty good to me.

14. Feb 26, 2008

### chocolatelover

Thank you very much

Regards

15. Feb 26, 2008

### chocolatelover

Could someone please check this? I want to make sure I'm doing it right.

Thank you

16. Feb 26, 2008

### chocokat

The first 16 should be x1, i.e. 3. Be careful of these little errors!

17. Feb 26, 2008

### chocolatelover

Thank you very much

What do you mean x1, i.e. 3? It should be multiplied by 1?

Thank you

18. Feb 26, 2008

### chocokat

This is the equation you are using:

What is the value of xi, and what is the value you used when you calculated it?
(I'm sorry, I used x1 above when I should have used xi)

Last edited: Feb 26, 2008
19. Feb 26, 2008

### chocolatelover

It's okay. It should have been 3, right? Does -26.25 look alright?

Last edited: Feb 26, 2008
20. Feb 26, 2008

### chocokat

Yeah, you've got it now.

21. Feb 26, 2008

### chocolatelover

Thank you very much

Regards