Position and Coordinate Systems

Every later chapter of this course makes statements about two things: the position of an object, and the change of that position with time. This chapter defines the terms for both. Every quantity is stated in SI units (meters, seconds, kilograms, and joules), so each number has exactly one unit.

Locating a point in the plane

The Position of an object is a vector from a fixed origin to the object. In the plane, the position has two coordinates: the Horizontal coordinate and the Vertical coordinate. Each coordinate is a function of Time. So the position at time $\explain{motion.time}{t}$ is the pair of the two coordinates at time $\explain{motion.time}{t}$:

$$\explain{motion.position}{\mathbf{r}}(\explain{motion.time}{t}) = \bigl(\explain{coordinate-x}{x}(\explain{motion.time}{t}),\ \explain{coordinate-y}{y}(\explain{motion.time}{t})\bigr)$$

(1.1.1)

The two coordinates in (1.1.1) are separate functions of the same time $\explain{motion.time}{t}$. So the motion of a projectile can be analyzed as two one-dimensional motions, one horizontal and one vertical, with one common time [1].

Two ways to name the same point

The horizontal and the vertical coordinate are the Cartesian coordinates of a point. Polar coordinates are a second way to state the same point. They give the distance of the point from the origin, which is the Polar radius, and the direction of the point from the origin, which is the Polar angle. The two coordinate systems contain the same information, so each converts into the other.

From Cartesian to polar

The Pythagorean theorem gives the polar radius from the two Cartesian coordinates, and the arctangent gives the polar angle:

$$\explain{polar-radius}{\rho} = \sqrt{\explain{coordinate-x}{x}^{2} + \explain{coordinate-y}{y}^{2}}, \qquad \explain{polar-angle}{\varphi} = \arctan\!\left(\frac{\explain{coordinate-y}{y}}{\explain{coordinate-x}{x}}\right)$$

(1.1.2)

The arctangent in (1.1.2) gives angles between $-90^{\circ}$ and $90^{\circ}$ only. These angles cover the right half of the plane. For a point in the left half of the plane, add $180^{\circ}$ to the result. For example, the points $(-1, -1)$ and $(1, 1)$ have the same ratio $\explain{coordinate-y}{y}/\explain{coordinate-x}{x}$, but they are on opposite sides of the origin. The conversion from polar to Cartesian coordinates in § 1.1.2.2 has no such restriction.

From polar to Cartesian

The polar radius multiplied by the cosine and by the sine of the polar angle gives the two Cartesian coordinates:

$$\explain{coordinate-x}{x} = \explain{polar-radius}{\rho}\cos(\explain{polar-angle}{\varphi}), \qquad \explain{coordinate-y}{y} = \explain{polar-radius}{\rho}\sin(\explain{polar-angle}{\varphi})$$

(1.1.3)

A launch velocity is usually stated in polar form: a speed and an angle above the horizontal. The conversion in (1.1.3) applies to every vector in the plane. So it also gives the horizontal and the vertical component of a launch velocity.

Velocity and acceleration as derivatives

The Velocity vector is defined as the time derivative of the position. The derivative of a vector is the vector of the derivatives of its coordinates:

$$\explain{velocity-vector}{\mathbf{v}}(\explain{motion.time}{t}) = \frac{d\explain{motion.position}{\mathbf{r}}}{d\explain{motion.time}{t}} = \left(\frac{d\explain{coordinate-x}{x}}{d\explain{motion.time}{t}},\ \frac{d\explain{coordinate-y}{y}}{d\explain{motion.time}{t}}\right)$$

(1.1.4)

A dot over a letter is a short way to write the Time derivative: $\explain{motion.time-derivative}{\dot{\explain{coordinate-x}{x}}}$ is the time derivative of the horizontal coordinate. With this notation, (1.1.4) is $\explain{velocity-vector}{\mathbf{v}}(\explain{motion.time}{t}) = (\explain{motion.time-derivative}{\dot{\explain{coordinate-x}{x}}},\ \explain{motion.time-derivative}{\dot{\explain{coordinate-y}{y}}})$.

The Acceleration is defined as the time derivative of the velocity. So the acceleration is the second time derivative of the position:

$$\explain{motion.acceleration}{\mathbf{a}}(\explain{motion.time}{t}) = \frac{d\explain{velocity-vector}{\mathbf{v}}}{d\explain{motion.time}{t}} = \frac{d^{2}\explain{motion.position}{\mathbf{r}}}{d\explain{motion.time}{t}^{2}}$$

(1.1.5)

If the position is twice differentiable with respect to time, (1.1.4) and (1.1.5) are correct. Neither equation assumes that the acceleration is constant in size or in direction.

The special case: constant acceleration

Every later scenario in this course adds one assumption: the Acceleration is constant in size and in direction.

Integrating once: velocity

With a constant acceleration, integrating (1.1.5) once with respect to time gives the velocity at time $\explain{motion.time}{t}$ from the initial velocity:

$$\explain{velocity-vector}{\mathbf{v}}(\explain{motion.time}{t}) = \explain{velocity-vector}{\mathbf{v}}_0 + \explain{motion.acceleration}{\mathbf{a}}\,\explain{motion.time}{t}$$

(1.1.6)

Integrating twice: position

Integrating (1.1.6) once more with respect to time gives the position at time $\explain{motion.time}{t}$ from the initial position and the initial velocity:

$$\explain{motion.position}{\mathbf{r}}(\explain{motion.time}{t}) = \explain{motion.position}{\mathbf{r}}_0 + \explain{velocity-vector}{\mathbf{v}}_0\,\explain{motion.time}{t} + \tfrac{1}{2}\explain{motion.acceleration}{\mathbf{a}}\,\explain{motion.time}{t}^{2}$$

(1.1.7)

The later chapters of this course use (1.1.7) more than any other equation. Its horizontal and its vertical coordinate give two independent equations. For a projectile, gravity accelerates only the vertical motion. So the horizontal velocity component is constant. The vertical velocity component decreases, is zero at the highest point, and is negative after the highest point.

Example 1.1.1 Calculation examples between coordinate systems

Cartesian to polar

A point is at $\explain{coordinate-x}{x} = 3\ \text{m}$ and $\explain{coordinate-y}{y} = 4\ \text{m}$. Substituting these coordinates into (1.1.2) gives its Polar radius:

$$\begin{aligned} \explain{polar-radius}{\rho} &= \sqrt{(3\ \text{m})^{2} + (4\ \text{m})^{2}}\\ &= \sqrt{9\ \text{m}^2 + 16\ \text{m}^2}\\ &= 5\ \text{m} \end{aligned}$$

The point is in the right half of the plane, so its Polar angle is $\arctan(4/3) \approx 53.1^{\circ}$ above the horizontal (approximately $0.93\ \text{rad}$).

Polar to Cartesian

A ball is thrown with a speed of $10\ \text{m/s}$ at $30^{\circ}$ above the horizontal. Applying (1.1.3) to this launch velocity gives its horizontal and vertical components:

$$\begin{aligned} \explain{velocity-vector}{\mathbf{v}}_0 &= \bigl(10\ \text{m/s}\cos(30^{\circ}),\ 10\ \text{m/s}\sin(30^{\circ})\bigr)\\ &\approx (8.66,\ 5.00)\ \text{m/s} \end{aligned}$$

Table 1.1.1Two points in both coordinate systems. Points A and B, each with its Cartesian coordinates x and y and its polar coordinates ρ and φ.

Point	Horizontal coordinate (meters)$\explain{coordinate-x}{x}\ \text{(m)}$	Vertical coordinate (meters)$\explain{coordinate-y}{y}\ \text{(m)}$	Polar radius (meters)$\explain{polar-radius}{\rho}\ \text{(m)}$	Polar angle (degrees)$\explain{polar-angle}{\varphi}\ \text{(°)}$
A	3	4	5	53.1
B	8.66	5.00	10	30.0

Table 1.1.1 states two points in both coordinate systems. Point A is the point of the first example in Example 1.1.1. Point B has the numbers of the second example, as a position in $\text{m}$ instead of a velocity in $\text{m/s}$.

Knowledge check 1.1.1 Position and coordinates

1. DirectQuestion 1Not answered

2. DirectQuestion 2Not answered

References

1. Ling et al., University Physics Volume 1 (2016). ch. 3, “Time, Position, and Displacement”. https://openstax.org/details/books/university-physics-volume-1 draft ↩