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$\text{Slope= tan(angle with respect to positive X-axis)= scalar output}$

$\text{velocity= a vector }$

graph

Source: Hugh D Young_ Roger A Freedman - University Physics with Modern Physics In SI Units (2019, Pearson) Page-67

Then, doubts are;

1.Is the relation "Slope of tangent=instantaneous x-velocity" valid, as it would mean "scalar=vector"?

2.Even if I write "Slope of tangent=instantaneous x-speed", if the tangent makes obtuse angle slope will be negative, and we know that instantaneous speed is magnitude of instantaneous velocity, which makes instantaneous speed a positive term. So, what exactly does the slope give?

Extra information:enter image description here

Source: Hugh D Young_ Roger A Freedman - University Physics with Modern Physics In SI Units (2019, Pearson) Page-67

Similar confusion arises in,$\text{ "Area under a x-t graph = change in x-velocity from time 0 to time t"}$, with right hand side vector and left hand side (area) as scalar. I think answer to the original question, provides solution to this confusion as well.

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Keep in mind there is a difference between a vector and a component of a vector. The velocity is a vector, the x-component of the velocity (or the "$x$-velocity" in the language your book uses) is just a component of a vector, which is a scalar.

The velocity vector can be expanded in terms of unit vectors $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$ (which satisfy $\mathbf{e}_x\cdot \mathbf{e}_x=1, \mathbf{e}_x\cdot \mathbf{e}_y=0$, etc): \begin{equation} \mathbf{v}=v_x \mathbf{e}_x + v_y \mathbf{e}_y + v_z \mathbf{e}_z \end{equation} The slope of the line you wrote down gives you $v_x$, which is a scalar given by $v_x=\mathbf{v}\cdot \mathbf{e}_x$. This quantity is a scalar, so there's no problem setting it equal to a slope. It is also a signed quantity, so it's not a speed (there's no requirement that $v_x>0$).

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  • $\begingroup$ A component of a vector is a vector in its own right. Otherwise, you couldn't add vector components to arrive at a resultant. $\endgroup$
    – David White
    Commented Feb 11, 2021 at 1:34
  • $\begingroup$ @DavidWhite $v_x=\mathbf{v}\cdot{e}_x$ is a scalar, not a vector, and this is the quantity I mean by "the $x$ component of $\mathbf{v}$" (I also think this is standard usage). Of course $v_x \mathbf{e}_x$ is a vector but I would not refer to this as the $x$ component of $\mathbf{v}$. $\endgroup$
    – Andrew
    Commented Feb 11, 2021 at 1:38
  • $\begingroup$ @DavidWhite "CAUTION: Components are not vectors", I just copy-pasted from the same book mentioned in question (Page 42). In general "component of a vector" means "scalar components" and "vector components(needs to explicitly mentioned) implies "vector components", this is the term you have mentioned as "A component of a vector..." $\endgroup$
    – Sahil
    Commented Feb 11, 2021 at 2:21
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    $\begingroup$ @Sahil, if the vector component has a magnitude (it surely does) and if the vector component has a direction (it should), it qualifies as a vector. If you are just talking about the magnitude of a vector component, that is a scalar. If that quote is from your book, then the book's authors are not using the word "component" the same way that I used it during the 13 years that I was teaching physics to high school students, and they are not using that word the same way that I have seen it used in every other physics book that I have read or used. $\endgroup$
    – David White
    Commented Feb 11, 2021 at 2:59
  • $\begingroup$ @DavidWhite, From what I have learnt using the book, Magnitude is a positive entity, scalar on the other hand can be positive or negative. Vector component is a vector, scalar component(generally termed as "component") is a scalar. $\endgroup$
    – Sahil
    Commented Feb 11, 2021 at 3:06

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