P01-280 – Relative Motion in 2D

Relative Motion in 1D and 2D

In this short part, we only seek to introduce the idea of relative motion without overly complicating the examples nor taking the concept as far as possible. We will simply focus on two frames of reference A and B that are moving relative to each other at a constant velocity, first in one dimension, then in two dimensions.

One-dimensional relative motion: position

Consider a passenger sitting in a train that is traveling with an increasing speed (equal to $3\ m/s$ in the figure) as it pulls out of the station. Reference frame A has its origin on the platform and is at rest with respect to the ground while reference frame B has its origin on the floor of the train car and is therefore traveling at a speed of $3\ m/s$ relative to the ground.

We let the $x$-position of the passenger, as measured in reference frame B, be denoted by $x_{P/B}$ and we let the $x$-position of the origin of reference frame B, as measured in reference frame A, be denoted by $x_{B/A}$.

We can then notice that the $x$-position of the passenger as measured in reference frame A is given by

\boxed{x_{P/A}=x_{B/A}+x_{P/B}}

where $x_{P/A}$ reads ”$x$ of passenger P relative to frame A” and denotes the $x$-position of the passenger as measured in reference frame A. With the positions indicated in the figure above, we then find that

x_{P/A}=10+3=13\ m

Note: the position $x_{P/B}$ of the passenger with respect to reference frame B is constant because the frame moves with the train car. However, the position $x_{P/A}$ of the passenger with respect to reference frame A changes with time because the train is moving relative to the ground.

One-dimensional relative motion: velocity and acceleration

If we consider the expression for $x_{P/A}$ found above and we take the derivative of this expression with respect to time, we find that the speed of the passenger, as measured in reference frame A, is given by

\frac{d}{dt}\left(x_{P/A}\right)=\frac{d}{dt}\left(x_{B/A}\right)+\frac{d}{dt}\left(x_{P/B}\right)\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{v_{P/A}=v_{B/A}+v_{P/B}}

In this particular example, since the passenger is sitting and is therefore at rest with respect to reference frame B, we have $v_{P/B}=0$ and therefore

v_{P/A}=v_{B/A}

This indicates that both the passager and reference frame B have the same speed in reference frame A which makes sense because both are moving with the train car but not relative to each other.

However, if we now assume that the passenger is walking toward the right end of the train car with a speed $v_{P/B}=2\ m/s$ measured with respect to the train car, we find that his speed relative to reference frame A (at the moment corresponding to the above figure) is equal to

v_{P/A}=v_{B/A}+v_{P/B}=3+2=5\ m/s

This indicates that, relative to the ground, the passenger is traveling at a speed of $5\ m/s$ which makes sense because the combined effect of the train car moving and the passenger moving with respect to the train car then gives the impression that the passenger is moving faster with respect to the ground.

Assuming that the passenger is still walking toward the right end of the car with speed $v_{P/B}=2\ m/s$, we take the derivative with respect to time of the expression found above for $v_{P/A}$, we find that the acceleration of the passenger, as measured in reference frame A, is equal to

\frac{d}{dt}\left(v_{P/A}\right)=\frac{d}{dt}\left(v_{B/A}\right)+\frac{d}{dt}\left(v_{P/B}\right)\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{a_{P/A}=a_{B/A}+a_{P/B}}

We initially assumed that the speed of the train is increasing and therefore $a_{B/A}\ne 0$. However, if the passenger walks at constant speed relative to the train, then $a_{P/B}=0\ \ m/s^2$ i.e. we have

a_{P/A}=a_{B/A}

Note: if the passenger started walking faster and faster, then we would have $a_{P/B}\ne 0$ and both accelerations would add to yield $a_{P/A}$.

Two-dimensional relative motion

Two-dimensional relative motion: position

In this paragraph, we extend the idea to relative motion along $x$ and $y$ simultaneously. Consider a student riding an escalator and tossing a tennis ball up and down as she climbs.

Reference frame A has its origin on the ground (lower level) and is at rest with respect to the ground while reference frame B has its origin on the stair of the escalator the student is standing on and is therefore traveling at a velocity of ${\overrightarrow{v}}_{B/A}$ relative to the ground.

We let ${\overrightarrow{r}}_{B/A}$ denote the position vector of the origin of reference B as measured with respect to reference frame A. Similarly, we let ${\overrightarrow{r}}_{T/B}$ denote the position vector of the tennis ball as measured with respect to reference frame B.

The position vector of the tennis ball in reference frame A is then given by

\boxed{{\overrightarrow{r}}_{T/A}={\overrightarrow{r}}_{B/A}+{\overrightarrow{r}}_{T/B}}

which can be written in component form as follows

\boxed{{\overrightarrow{r}}_{T/A}=\left(x_{B/A}+x_{T/B}\right)\hat{x}+\left(y_{B/A}+y_{T/B}\right)\hat{y}}

This form allows us to see that we are simply generalizing the concept of one-dimensional motion to two dimensions by adding a $y$-component.

In then follows that, by taking the derivative of the above expression with respect to time we find that the velocity of the tennis ball measured in reference frame A is given by

\boxed{{\overrightarrow{v}}_{T/A}={\overrightarrow{v}}_{B/A}+{\overrightarrow{v}}_{T/B}}

Similarly, by taking the derivative with respect to time yet again, we find that the acceleration of the tennis ball measured in reference frame A is given by

\boxed{{\overrightarrow{a}}_{T/A}={\overrightarrow{a}}_{B/A}+{\overrightarrow{a}}_{T/B}}

Note: if the velocity of the escalator is constant, then we can conclude that $a_{B/A}=0$.