Why Use Unit Vectors in Calculations?

[vetgirl1990]

I'm confused about what situations you should use unit vectors in... and it seems that when I approach the same problem using unit vectors vs. without unit vectors, I get different answers. Why?
To illustrate my confusion, here's an example that I tried solving using unit vectors, and without unit vectors.

For example:

Homework Statement

A particle moves in the xy plane from the origin at t=0 with an initial velocity having an x-component of 20m/s, and a y-component of -15m/s. The particle has an acceleration in the x-direction of a_x=4m/s²

What is the velocity of the particle at 5 seconds?

2. Homework Equations
Velocity as a function of time: v_f=v_i+at

The Attempt at a Solution

Expressed as unit vectors:
v_f = (v_ix+a_xt)i + (v_iy+a_yt)j
v_f = (20+4t)i - 15j
v_f = [20 +4(5)]i - 15j = (40i - 15j)m/s
Expressed into x and y components: v_fx=40m/s, v_fy=-15m/s
v_f = sqrt(40² + (-15)²) = 37m/s

Expressed as "regular" vectors:
x direction: v_fx = v_ix+a_xt = 20+4t
y direction: v_fy = v_iy+a_yt = -15 +4(0) = -15
v_f = sqrt[(20+4t)² + (-15)²] = sqrt(625 + 160t + 16t²) = 43m/s

EDIT: I rechecked my calculations, and realized that my final answer for the unit vector approach is also 43m/s... D'oh!
Given my mistake, my question is now: Will a unit vector approach always give me the same answer as when I don't use unit vectors?

 

[SteamKing]

I'm confused about what situations you should use unit vectors in... and it seems that when I approach the same problem using unit vectors vs. without unit vectors, I get different answers. Why?
To illustrate my confusion, here's an example that I tried solving using unit vectors, and without unit vectors.

For example:

Homework Statement

A particle moves in the xy plane from the origin at t=0 with an initial velocity having an x-component of 20m/s, and a y-component of -15m/s. The particle has an acceleration in the x-direction of a_x=4m/s²

What is the velocity of the particle at 5 seconds?

2. Homework Equations
Velocity as a function of time: v_f=v_i+at

The Attempt at a Solution

Expressed as unit vectors:
v_f = (v_ix+a_xt)i + (v_iy+a_yt)j
v_f = (20+4t)i - 15j
v_f = [20 +4(5)]i - 15j = (40i - 15j)m/s
Expressed into x and y components: v_fx=40m/s, v_fy=-15m/s
v_f = sqrt(40² + (-15)²) = 37m/s

Expressed as "regular" vectors:
x direction: v_fx = v_ix+a_xt = 20+4t
y direction: v_fy = v_iy+a_yt = -15 +4(0) = -15
v_f = sqrt[(20+4t)² + (-15)²] = sqrt(625 + 160t + 16t²) = 43m/s

In the last line of the 'unit vectors' calculation, check your arithmetic when you calculate the velocity at t = 5 sec.

In the last line of your 'regular' calculation, why did you expand (20 + 4t)² ? Why didn't you just evaluate (20 + 4t) and then square the result?

 

[vetgirl1990]

In the last line of the 'unit vectors' calculation, check your arithmetic when you calculate the velocity at t = 5 sec.

In the last line of your 'regular' calculation, why did you expand (20 + 4t)² ? Why didn't you just evaluate (20 + 4t) and then square the result?

Such a trivial mistake on my part! Yes, I redid my calculations and got 43m/s for both approaches. Now, will a "unit vector" approach always give me the same answer as when I don't use unit vectors?

 

[SteamKing]

Such a trivial mistake on my part! Yes, I redid my calculations and got 43m/s for both approaches. Now, will a "unit vector" approach always give me the same answer as when I don't use unit vectors?

It should. You still have to keep your arithmetic checked, though.

 

[Mister T]

Will a unit vector approach always give me the same answer as when I don't use unit vectors?

Of course. It's just a different way of writing things out. Note that you can write a vector equation with unit vectors, something you cannot do without them.

For example, $\vec{J}=25 \hat i+7 \hat j$