Tutorials › Calculus III › Relative Position and Velocity

Home  /  Tutorials  /  Motion Along Space Curves

Motion Along Space Curves · Tutorial 163 of 999

Relative Position and Velocity

Use vector differences to describe how two objects are positioned and moving relative to one another, then test for separation and interception.

Intermediate 9 min read

What You'll Learn

  • Define relative position and relative velocity using a consistent choice of reference object.
  • Calculate both quantities from position functions and interpret their directions.
  • Find the distance between two moving objects at a specified time.
  • Test interception by solving all position-coordinate equations at the same time.
  • Distinguish a close pass from an actual interception.
  • Check relative-motion calculations by comparing the original position functions.

Comparing Two Motions

When two objects move through space, it is often more useful to compare their motions than to describe each motion separately. A vehicle may be moving east, for example, while another vehicle is moving west. Their individual velocities describe their motion relative to the chosen coordinate system; their relative velocity describes how quickly and in what direction their separation vector is changing.

The key operation is subtraction. Subtract the position of the object used as the reference from the position of the other object. Apply the same subtraction to their velocity vectors. The order matters: reversing the order reverses both relative vectors.

Definition: Let objects \(A\) and \(B\) have position functions \(\mathbf r_A(t)\) and \(\mathbf r_B(t)\), defined at the same time \(t\). The position of \(B\) relative to \(A\) is \(\mathbf r_{B/A}(t)=\mathbf r_B(t)-\mathbf r_A(t)\). If their velocities are \(\mathbf v_A(t)\) and \(\mathbf v_B(t)\), the velocity of \(B\) relative to \(A\) is \(\mathbf v_{B/A}(t)=\mathbf v_B(t)-\mathbf v_A(t)\).

The subscript \(B/A\) means “\(B\) relative to \(A\).” The vector \(\mathbf r_{B/A}\) points from \(A\) to \(B\). Its magnitude is the distance between them. The vector \(\mathbf v_{B/A}\) describes the change in that relative position per unit time. If positions are measured in metres and time in seconds, relative position is measured in metres and relative velocity in metres per second.

For position functions that are differentiable, the velocity of each object is the derivative of its position. Subtracting the positions and differentiating gives the relative-velocity formula. This is the reason the same subtraction rule works for both quantities.

Why the Relative-Velocity Formula Works

1
Form the relative position.
By definition, \(\mathbf r_{B/A}(t)=\mathbf r_B(t)-\mathbf r_A(t)\).
2
Differentiate each component.
When both position functions are differentiable, the derivative of a difference is the difference of the derivatives.
3
Identify the velocities.
Since \(\mathbf r_A'(t)=\mathbf v_A(t)\) and \(\mathbf r_B'(t)=\mathbf v_B(t)\), the derivative is \(\mathbf v_B(t)-\mathbf v_A(t)\).
$$ \frac{d}{dt}\mathbf r_{B/A}(t) = \frac{d}{dt}\bigl(\mathbf r_B(t)-\mathbf r_A(t)\bigr) = \mathbf v_B(t)-\mathbf v_A(t) = \mathbf v_{B/A}(t). $$

This derivation uses only the componentwise derivative rule for a difference. It also shows why relative velocity is not found by adding the two velocity vectors: the relevant change is the difference between their motions.

Calculating Relative Position and Separation

To find the separation at a particular time, first evaluate both position functions at that same time. Subtract in the stated order to get the relative-position vector. Then take its magnitude. In two or three dimensions, the magnitude is the square root of the sum of the squares of its components.

Worked Example: Relative Motion at a Given Time

Suppose two objects move in a plane, with positions in metres at time \(t\) seconds given by

$$ \mathbf r_A(t)=\langle 1+2t,\;3-t\rangle, \qquad \mathbf r_B(t)=\langle 7-t,\;-1+2t\rangle. $$

Find the relative position and relative velocity of \(B\) with respect to \(A\), and their separation at \(t=1\).

Subtract the position of \(A\) from the position of \(B\), component by component:

$$ \begin{aligned} \mathbf r_{B/A}(t) &=\langle 7-t,\;-1+2t\rangle-\langle 1+2t,\;3-t\rangle\\ &=\langle 7-t-(1+2t),\;-1+2t-(3-t)\rangle\\ &=\langle 6-3t,\;-4+3t\rangle. \end{aligned} $$

Differentiate the components to find the velocities. Equivalently, differentiate the relative-position function:

$$ \mathbf v_A(t)=\langle 2,-1\rangle,\qquad \mathbf v_B(t)=\langle -1,2\rangle,\qquad \mathbf v_{B/A}(t)=\langle -3,3\rangle\ \text{m/s}. $$

At \(t=1\) second, the relative position is

$$ \mathbf r_{B/A}(1)=\langle 6-3,\;-4+3\rangle =\langle 3,-1\rangle\ \text{m}. $$

Thus the separation is \(\sqrt{3^2+(-1)^2}=\sqrt{10}\) metres, approximately \(3.16\) metres to three significant figures. Check directly: \(\mathbf r_A(1)=\langle 3,2\rangle\) and \(\mathbf r_B(1)=\langle 6,1\rangle\); subtracting gives \(\langle 3,-1\rangle\), with the same magnitude.

Interception Requires Simultaneous Equality

An interception occurs only if the objects occupy the same point at the same time. In relative-position language, that means the relative-position vector is the zero vector. For functions written in coordinates, every component must equal zero for one and the same time. Solving only one coordinate equation is not enough.

Interception condition: Objects \(A\) and \(B\) intercept at a time \(t_*\) in their shared time domain exactly when \(\mathbf r_A(t_*)=\mathbf r_B(t_*)\), or equivalently \(\mathbf r_{B/A}(t_*)=\mathbf 0\). For a physical motion restricted to \(t\geq 0\), the solution must also satisfy \(t_*\geq 0\).

Worked Example: Testing for an Interception

Two objects move in a plane, with positions in metres for \(t\geq 0\):

$$ \mathbf r_A(t)=\langle 3t,t\rangle, \qquad \mathbf r_B(t)=\langle 10+t,5\rangle. $$

Find whether \(B\) intercepts \(A\), and if so, when and where. The relative position is

$$ \begin{aligned} \mathbf r_{B/A}(t) &=\langle 10+t,5\rangle-\langle 3t,t\rangle\\ &=\langle 10-2t,5-t\rangle. \end{aligned} $$

For interception, both components must be zero. The first gives \(10-2t=0\), so \(t=5\). The second gives \(5-t=0\), also \(t=5\). This is a valid time in the domain, so interception occurs at \(t=5\) seconds. Substitution into either position function gives

$$ \mathbf r_A(5)=\langle 15,5\rangle\ \text{m}, \qquad \mathbf r_B(5)=\langle 15,5\rangle\ \text{m}. $$

The positions agree, confirming the interception. Their velocities are \(\mathbf v_A=\langle 3,1\rangle\) m/s and \(\mathbf v_B=\langle 1,0\rangle\) m/s, so \(\mathbf v_{B/A}=\langle -2,-1\rangle\) m/s. The objects meet, but their velocities are different; interception does not require them to travel together afterward.

A Close Pass Is Not Necessarily an Interception

When the relative-position vector is nonzero, the objects are separated. They may nevertheless pass close to one another. The magnitude of the relative-position vector measures that separation at each time, while a zero relative-position vector identifies an actual meeting. A small distance alone is not evidence of interception.

Worked Example: A Close Pass Without Interception

For \(t\geq 0\), let the positions, in metres, be

$$ \mathbf r_A(t)=\langle t,1\rangle, \qquad \mathbf r_B(t)=\langle 0,t\rangle. $$

The relative position and relative velocity of \(B\) with respect to \(A\) are

$$ \mathbf r_{B/A}(t)=\langle -t,t-1\rangle, \qquad \mathbf v_{B/A}(t)=\langle -1,1\rangle\ \text{m/s}. $$

For interception, the first component requires \(t=0\), while the second requires \(t=1\). No single time satisfies both equations, so there is no interception. To see how close they come, calculate the squared separation:

$$ \begin{aligned} \|\mathbf r_{B/A}(t)\|^2 &=(-t)^2+(t-1)^2\\ &=t^2+t^2-2t+1\\ &=2\left(t-\frac12\right)^2+\frac12. \end{aligned} $$

The squared term is nonnegative and is zero at \(t=\frac12\). Therefore the smallest possible squared separation is \(\frac12\) square metres, and the minimum separation is \(\sqrt{\frac12}=\frac{1}{\sqrt2}\) metres, approximately \(0.71\) metres to two decimal places. Check at \(t=\frac12\): the positions are \(\mathbf r_A=\langle \frac12,1\rangle\) and \(\mathbf r_B=\langle 0,\frac12\rangle\). Their difference is \(\langle-\frac12,-\frac12\rangle\), whose magnitude is indeed \(\sqrt{\frac14+\frac14}=\frac{1}{\sqrt2}\) metres, not zero.

Keep the Reference Order Consistent

The most frequent error is reversing the subtraction in one calculation but not the other. If position is defined as \(B\) relative to \(A\), use \(\mathbf r_B-\mathbf r_A\). The matching relative velocity is \(\mathbf v_B-\mathbf v_A\). Reversing both gives a valid description from \(A\) relative to \(B\), but mixing the orders gives inconsistent directions.

For example, in the first worked example, at \(t=1\) the correct relative position of \(B\) with respect to \(A\) is \(\langle 3,-1\rangle\) m. Reversing the subtraction gives \(\mathbf r_A(1)-\mathbf r_B(1)=\langle-3,1\rangle\) m. This is the vector from \(B\) to \(A\), not from \(A\) to \(B\). Its magnitude is unchanged, but its direction is reversed. A related mistake is to decide that an interception occurs because one component of relative position is zero. In the close-pass example, one component vanishes at \(t=0\) and the other at \(t=1\); since the times differ, the objects do not meet.

These comparisons are useful whenever the question concerns the gap between two moving objects: their distance at a stated time, whether they meet, or whether they merely pass nearby. The coordinate-system positions may be complicated, but their difference directly describes the motion that matters for those questions.

Check Your Understanding

Use the reference order stated in each question, and show the coordinate equations when testing for interception.

  1. If \(\mathbf r_A(t)=\langle 2t,1\rangle\) and \(\mathbf r_B(t)=\langle 5-t,4\rangle\), find \(\mathbf r_{B/A}(t)\) and \(\mathbf v_{B/A}(t)\).
  2. How is the separation of two objects obtained from their relative-position vector?
  3. What condition must all components of the relative-position vector satisfy for an interception?
  4. Can two objects have the same first coordinate at a time and still fail to intercept? Explain.
  5. If the relative position is \(\langle 4,-3\rangle\) metres, what is the separation?

Previous

Next

Back to Tutorials