Question: I'm currently working on a problem in which an object's speed remains the same, but it experiences a change in direction after some time. I'm fairly sure it experiences an acceleration in order to change direction.

However, I am confused because of the change in direction. 

I want to use the formula $A\_x = \Delta v / \Delta t $. Since the object experiences a change in direction at some point but still travels at the same speed, would the second velocity technically be negative? Like so:

$V\_1$ = 5.0m/s
,$V\_2$ = 5.0m/s,
$T\_1$ = 0 seconds, $T\_2$ = 10 seconds

$A\_x = V\_1 - (-V\_2) / T\_1 - T\_2 $

Also, does it matter which direction? Would it be the same as travelling right and then turning left/right/around?

Answer: 
> 
> I'm fairly sure it experiences an acceleration in order to change direction.
> 
> 
> 

Yes, any change in velocity means an acceleration has occurred. That includes a change in direction of the velocity.

> 
> Since the object experiences a change in direction at some point but still travels at the same speed, would the second velocity technically be negative?
> 
> 
> 

If the first velocity is in the positive direction, yes the second might be negative. Is this a 1-dimensional problem?

> 
> I want to use the formula $A\_x=\Delta v/ \Delta t.$
> 
> 
> 

That will give you the **average** acceleration over the total time. It tells you nothing about the instantaneous acceleration at any specific time. Also, for $A\_x$ you would use $\Delta v\_x$.

Any curvilinear motion (which can't happen in 1-D motion) requires a sideways acceleration component of $$ a\_{\perp} = \frac{v^2}{r} $$ where $v$ is the instantaneous speed and $r$ is the instantaneous radius of curvature.
You've confused velocity and speed. Speed is independent of direction, and is therefor always positive. For a 2-dimensional analysis, you have to consider the two different components of the velocity, Vx and Vy, but the speed S is $$S=\sqrt{V\_x^2 + V\_y^2}$$

Let's take a version of your example. An object travels along the x-axis at 5 m/s in the positive direction. 10 seconds later it is seen to be travelling parallel to the y-axis in the positive (y) direction. In the x-axis, the change in velocity was $$\Delta V\_x = 0 - 5 = -5 m/s$$ and for the y-axis, $$\Delta V\_y = 5 - 0 = 5 m/s$$ So the average acceleration over the 10 seconds for each axis is $$a\_x =\frac {Delta V\_x}{Delta t} = \frac {-5}{10} = -0.5 \frac{m}{s^2}$$ and $$a\_y =\frac {Delta V\_y}{Delta t} = \frac {5}{10} = 0.5 \frac{m}{s^2}$$

If the object had been given a negative y-velocity, the y-acceleration would have been negative.

While this is pretty obvious for right-angles, let's say the object changes course by 45 degrees, going up and to the right. Then the velocities are equal (since it's 45 degrees), and since the speed remains the same, $$5=\sqrt{V\_x^2 + V\_y^2} = \sqrt{2V\_x^2}$$ and $$V\_x = \frac{5}{sqrt{2}} = \frac{5}{1.414} = 3.54 \frac{m}{s}$$ and Vy is the same. You can now calculate the accelerations required.