<< Chapter < Page | Chapter >> Page > |
$$\begin{array}{l}{v}_{CA}=\frac{\u0111{x}_{CA}}{\u0111t}\end{array}$$
and
$$\begin{array}{l}{v}_{CB}=\frac{\u0111{x}_{CB}}{\u0111t}\end{array}$$
Now, we can obtain relation between these two velocities, using the relation ${x}_{CA}={x}_{BA}+{x}_{CB}$ and differentiating the terms of the equation with respect to time :
$$\begin{array}{l}\frac{\u0111{x}_{CA}}{\u0111t}=\frac{\u0111{x}_{BA}}{\u0111t}+\frac{\u0111{x}_{CB}}{\u0111t}\end{array}$$
$$\begin{array}{l}\Rightarrow {v}_{CA}={v}_{BA}+{v}_{CB}\end{array}$$
The meaning of the subscripted velocities are :
Problem : Two cars, standing a distance apart, start moving towards each other with speeds 1 m/s and 2 m/s along a straight road. What is the speed with which they approach each other ?
Solution : Let us consider that "A" denotes Earth, "B" denotes first car and "C" denotes second car. The equation of relative velocity for this case is :
$$\begin{array}{l}\Rightarrow {v}_{CA}={v}_{BA}+{v}_{CB}\end{array}$$
Here, we need to fix a reference direction to assign sign to the velocities as they are moving opposite to each other and should have opposite signs. Let us consider that the direction of the velocity of B is in the reference direction, then
$$\begin{array}{l}{v}_{BA}=1\phantom{\rule{2pt}{0ex}}m\mathrm{/}s\phantom{\rule{4pt}{0ex}}\mathrm{and}\phantom{\rule{4pt}{0ex}}{v}_{CA}=-2\phantom{\rule{2pt}{0ex}}m\mathrm{/}s.\end{array}$$
Now :
$$\begin{array}{l}{v}_{CA}={v}_{BA}+{v}_{CB}\end{array}$$
$$\begin{array}{l}\Rightarrow -2=1+{v}_{CB}\\ \Rightarrow {v}_{CB}=-2-1=-3\phantom{\rule{2pt}{0ex}}m\mathrm{/}s\end{array}$$
This means that the car "C" is approaching "B" with a speed of -3 m/s along the straight road. Equivalently, it means that the car "B" is approaching "C" with a speed of 3 m/s along the straight road. We, therefore, say that the two cars approach each other with a relative speed of 3 m/s.
If the object being observed is accelerated, then its acceleration is obtained by the time derivative of velocity. Differentiating equation of relative velocity, we have :
$$\begin{array}{l}{v}_{CA}={v}_{BA}+{v}_{CB}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{\u0111{v}_{CA}}{\u0111t}=\frac{\u0111{v}_{BA}}{\u0111t}+\frac{\u0111{v}_{CB}}{\u0111t}\end{array}$$
$$\begin{array}{l}\Rightarrow {a}_{CA}={a}_{BA}+{a}_{CB}\end{array}$$
The meaning of the subscripted accelerations are :
But we have restricted ourselves to reference systems which are moving at constant velocity. This means that relative velocity of "B" with respect to "A" is a constant. In other words, the acceleration of "B" with respect to "A" is zero i.e. ${a}_{BA}=0$ . Hence,
$$\begin{array}{l}\Rightarrow {a}_{CA}={a}_{CB}\end{array}$$
The observers moving at constant velocity, therefore, measure same acceleration of the object. As a matter of fact, this result is characteristics of inertial frame of reference. The reference frames, which measure same acceleration of an object, are inertial frames of reference.
The important aspect of relative velocity in one dimension is that velocity has only two possible directions. We need not use vector notation to write or evaluate equation of relative velocities in one dimension. The velocity, therefore, can be treated as signed scalar variable; plus sign indicating velocity in the reference direction and minus sign indicating velocity in opposite to the reference direction.
The equation of relative velocities refers velocities in relation to different reference system.
$$\begin{array}{l}{v}_{CA}={v}_{BA}+{v}_{CB}\end{array}$$
We note that two of the velocities are referred to A. In case, “A” denotes Earth’s reference, then we can conveniently drop the reference. A velocity without reference to any frame shall then mean Earth’s frame of reference.
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?