Determine the acceleration of the object during this motion

[blayman5]

Homework Statement

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.

Homework Equations

v=vi+at
V^2=Vi^2+2ax

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?

 

[LowlyPion]

Homework Statement

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.

Homework Equations

v=vi+at
V^2=Vi^2+2ax

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 wrongI 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?

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?

 

[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:

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

 

[blayman5]

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

 

[LowlyPion]

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

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

 

[chislam]

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

We gave you the 2 that you should use.

 

[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?

 

[LowlyPion]

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?

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.

 

[chislam]

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

Look closely at what you left out at this stage.

 

[blayman5]

Ah yes, i forgot the t.

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

 

[chislam]

Indeed that is correct.

 

[blayman5]

thanks a lot chislam and lowlypion