How did u subtract for V in the formula for workdone, I also do not understand how u squared the V

The velocity $\mathbf{v}$ is a vector. The speed $v$ is the magnitude of this vector. If $\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j} + v_z\mathbf{k}$, then the square of the speed is its magnitude squared:

$$v^2 = |\mathbf{v}|^2 = v_x^2 + v_y^2 + v_z^2$$

• At $t=1$, the velocity vector was $\mathbf{v}(1) = (4\mathbf{i} + 3\mathbf{j} + \mathbf{k})\text{ m s}^{-1}$. So, $v_1^2 = (4)^2 + (3)^2 + (1)^2 = 16 + 9 + 1 = 26$.
• At $t=2$, the velocity vector was $\mathbf{v}(2) = (4\mathbf{i} + 12\mathbf{j} + 2\mathbf{k})\text{ m s}^{-1}$. So, $v_2^2 = (4)^2 + (12)^2 + (2)^2 = 16 + 144 + 4 = 164$.

The formula for work done is $W = \frac{1}{2}m(v_2^2 - v_1^2)$. Here, $v_1^2$ and $v_2^2$ are scalar values (the squared speeds), not vectors.
We calculated $v_2^2 = 164$ and $v_1^2 = 26$.
The subtraction is simply:

$$v_2^2 - v_1^2 = 164 - 26 = 138$$

This scalar result is then used in the work done formula:

$$W = \frac{1}{2}(2)(138) = 138 \text{ J}$$

The subtraction is between the squared magnitudes of the velocities (which are scalar speeds), not a vector subtraction.