# Homework Help: Determine the acceleration of the object during this motion

1. Aug 31, 2008

### blayman5

1. The problem statement, all variables and given/known data

An object moving on the x axis with a constant acceleration increases its x coordinate
by 31 m in a time of 6.4 s and has a velocity
of 20 m/s at the end of this time.
Determine the acceleration of the object
during this motion.

2. Relevant equations
v=vi+at
V^2=Vi^2+2ax

3. The attempt at a solution
What I did:
found the initial velocity by dividing 31m/6.4s and got 4.84375
used galileo's formula: V^2=Vi^2+2ax
my answer was 6.3 or so and it was wrong

I am not so sure that the initial velocity is right, but I cannot think of any ideas. Can someone assist me in the right direction?

2. Aug 31, 2008

### LowlyPion

Your instinct is correct. The increase in distance on the constantly accelerating object won't be given by simply dividing the 2.

Rather don't you want to use the relationship that Vf = Vi + at?

3. Aug 31, 2008

### chislam

I'm assuming that LowlyPion is thinking the way that I am at the moment, so to add to his reply, I say:

Since you don't know the initial velocity, use the relationship that he provided and substitute it into the equation:

$$x = v_0t + \frac{1}{2}at^2$$

4. Aug 31, 2008

### blayman5

Ok, but the acceleration is also missing. Would I have to use one formula and substitute variables into another?

5. Aug 31, 2008

### LowlyPion

I think you are on to something. That would be the way to find a.

6. Sep 1, 2008

### chislam

We gave you the 2 that you should use.

7. Sep 1, 2008

### blayman5

alright, i substituted the vf=vi+at into x=vit+1/2at^2
vi=vf+at
x=(vf-at)+1/2at^2
x-vf=-at+1/2at^2
x-vf=at(-1+1/2t)
x-vf/((1/2t-1)t)=a
and i got around .7

right track?

8. Sep 1, 2008

### LowlyPion

I would suggest that you use the numbers of your problem, solving for unknowns along the way. It makes the algebra easier. (At least I find it so.)

In looking at your work I think, without scratching it out myself, that looks on the right road.

Last edited: Sep 1, 2008
9. Sep 1, 2008

### chislam

Look closely at what you left out at this stage.

10. Sep 1, 2008

### blayman5

Ah yes, i forgot the t.

x=(vf-at)t+1/2at^2
a=(-2(x/t-vf))/t
=4.736

11. Sep 1, 2008

### chislam

Indeed that is correct.

12. Sep 1, 2008

### blayman5

thanks a lot chislam and lowlypion