I was going through a textbook (Hibbeler, Engineering Mechanics: Dynamics 14th ed.) and the following passage stumped me. The chapter was on the relative motion of two particles using a translation of axes. I don't understand how the direction of $r_{B/A}$ stays constant between two moving objects or if I'm mis-interpreting what the passage is saying.

An equation that relates the velocities of the particles is determined by taking the time derivative of the above equation, i.e.

$v_B = v_A + v_{B/A}$ (12–34)

Here $v_B = dr_B/dt$ and $v_A = dr_A/dt$ refer to absolute velocities, since they are observed from the fixed frame; whereas the relative velocity $v_{B/A} = dr_{B/A}/dt$ is observed from the translating frame.

It is important to note that since the x', y', z' axes translate, the components of $r_{B/A}$ will not change direction and therefore the time derivative of these components will only have to account for the change in their magnitudes.

Equation 12–34 therefore states that the velocity of B is equal to the velocity of A plus (vectorially) the velocity of “B with respect to A,” as measured by the translating observer fixed in the x, y, z reference frame.


1 Answer 1


What they fail to mention is that the direction does not change in time ${\rm d}t$. Essentially since there are no rotations the direction of the relative position isn't rotating also.

Over finite time scales the direction does change. For me they are taking a simple observation and making it more complex than it needs.

  • $\begingroup$ I think I am mis-understanding what they mean by "components of $r_{A/B}$ will not change direction". I initially thought they mean the unit vector of $r_{A/B}$ has constant direction, which is probably a mis-interpretation. Do they mean that i, j, k (of the translating frame x',y',z'), won't change in direction relative to the fixed frame x,y,z? If so I'm missing the connection between that and your second point. $\endgroup$
    – Yandle
    Commented Jul 31, 2016 at 17:11

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