Is the relation "slope=velocity" mathematically valid?

Question

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

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?
Extra information:
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.

Andrew · Accepted Answer

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 $v_x>0$).

Is the relation "slope=velocity" mathematically valid?

One Answer

Add your own answers!

Ask a Question