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We have the first two equations of motion in $\mathrm{S.I}$ units as, $\vec{v} = [u(\vec{i}+\vec{j}+\vec{k}) + a(\vec{i}+\vec{j}+\vec{k})t] \space \mathrm{and} \space \vec{s} = [u(\vec{i}+\vec{j}+\vec{k}) t + \frac{a(\vec{i}+\vec{j}+\vec{k})t^2}{2}]$.

Where $u$, $v$, $a$ and $s$ are real numbers; $\vec{i}$, $\vec{j}$, $\vec{k}$ are unit vectors pointing in $x$, $y$, $z$ directions, respectively.

Can I write the third equation as, $\vec{v}\cdot\vec{v} - \vec{u}\cdot\vec{u} = 2\vec{a}\cdot\vec{s}$?

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    $\begingroup$ Your first two equations assume that $\mathbf u$ and $\mathbf a$ are colinear, which isn't generally true. By making that assumption you are implicitly restricting the system to one dimension. $\endgroup$
    – John Rennie
    Commented Jul 18, 2017 at 5:39
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    $\begingroup$ I should have mentioned that my study is along one direction. Given vectors in space, my first attempt would be to resolve them along three different directions and work with the equations above. $\endgroup$
    – R004
    Commented Jul 18, 2017 at 5:42
  • $\begingroup$ Let me modify them to suggest their existence in three dimensions. Give me a minute or two. $\endgroup$
    – R004
    Commented Jul 18, 2017 at 5:44
  • $\begingroup$ You cannot write the third equation in that way. I am assuming you are taking dot product or scalar product. So, there are no vectors left in your third equation as dot product of two vectors is a scalar $\endgroup$
    – Mriganka Parasar
    Commented Jul 18, 2017 at 5:56
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    $\begingroup$ The third equation is a $\mathrm{Scalar}$ equation. $\endgroup$
    – R004
    Commented Jul 18, 2017 at 5:58

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The velocity, acceleration and position vectors are defined in terms of each other as follows:

$$\vec{v} = \frac{d\vec{x}}{dt} \tag{1}$$

$$\vec{a} = \frac{d\vec{v}}{dt} \tag{2}$$

Using the previous two, you can obtain the third differential equation:

$$\vec{a} = \vec{v}\frac{d\vec{v}}{d\vec{x}} \tag{3}$$

We can rearrange the equations to obtain the following:

$$d\vec{x} = \vec{v} \space dt \tag{4}$$

$$d\vec{v} = \vec{a} \space dt \tag{5}$$

$$\vec{a}\space d\vec{x} = \vec{v} \space d\vec{v} \tag{6}$$

The equations $(1), (2), (3), (4), (5)$ and $(6)$ always hold true.

The equations of motion which you mentioned in your post hold true if and only if acceleration is constant. With that assumption, we can reduce the differential equations mentioned earlier to:

$$\vec{v} - \vec{v_0} = \vec{a} (t - t_0)$$

We generally label $\vec{v_0}$ as $\vec{u}$ and $t_0$ is taken as 0. From that we get:

$$\vec{v} - \vec{u} = \vec{a}t \implies \vec{v} = \vec{u} + \vec{a} t$$

Using the previous equation, we get:

$$d\vec{x} = \vec{v} dt = (\vec{u} + \vec{a}t)dt$$

$$\vec{x} - \vec{x_0} = \vec{u} (t - t_0) + \frac{a}{2}(t^2 - t_0^2)$$

We generally take $x_0$ and $t_0$ as $0$. Therefore, the aforementioned equationr educes to:

$$\vec{x} = \vec{u}t + \frac{a}{2}t^2$$

The last equation on integration yields:

$$\vec{a}.(\vec{x} - \vec{x_0}) = \frac{|\vec{v}|^2 - |\vec{u}|^2}{2} = \frac{\vec{v}.\vec{v} - \vec{u}.\vec{u}}{2}$$

We generally take $\vec{x_0}$ as $\vec{0}$. Therefore, the aforementioned equation reduces to:

$$2\vec{a}.\vec{x} = \vec{v}.\vec{v} - \vec{u}.\vec{u}$$

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    $\begingroup$ Let us consider the equation, $\vec{a} = \vec{v}\frac{d\vec{v}}{d\vec{x}} \tag{3}$. Here, why have you not considered any rules of vector multiplication between $\vec{v}$ and $\frac{d\vec{v}}{d\vec{x}} \tag{3}$? Also, the ratio of the two differentials suggests division of two vectors, does it not? $\endgroup$
    – R004
    Commented Jul 18, 2017 at 7:35
  • $\begingroup$ Integrating equation (6) mysteriously brought in the dot product, it seems. I'm quite curious to know your working with equations (3) and (6). If you used chain rule in eq. (2) to get (3), I have a doubt. The differentials $dx$ seem to cancel one another, so it's alright. But these differentials are vectors. Their division is not defined at least in the realm of undergraduate classical mechanics( I could be wrong ). $\endgroup$
    – R004
    Commented Jul 18, 2017 at 8:02
  • $\begingroup$ Another way to understand is as follows: $$\frac{d}{dt} v^2 = \frac{d}{dt} (\vec{v}.\vec{v}) = \vec{v}.\frac{d\vec{v}}{dt} + \frac{d\vec{v}}{dt}.\vec{v} = 2\vec{v}.\vec{a}$$ You can now integrate it to obtain the final result. $\endgroup$
    – Yashas
    Commented Jul 18, 2017 at 8:10

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