Physics Asked on May 9, 2021

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

$text{velocity= a vector }$

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?

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.

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 $mathbf{e}_xcdot mathbf{e}_x=1, mathbf{e}_xcdot mathbf{e}_y=0$, 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 $v_x>0$).

Correct answer by Andrew on May 9, 2021

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