Resultant and Relative Velocity

Welcome to one of the most practical and exciting areas of Mechanics! Have you ever looked out of a car window on the motorway and felt like a car overtaking you was moving at a crawling pace? Or have you watched a plane land in strong crosswinds, appearing to fly sideways towards the runway? These everyday phenomena are governed by resultant velocity and relative velocity.

In this chapter, we will master the tools needed to analyse motion from different perspectives, calculate true paths across windy skies or flowing rivers, and solve classic problems such as navigation, closest approach, and interception. Don't worry if vectors have felt a bit abstract in the past—we will break down every concept step by step!


1. Vectors Recap & Resultant Velocity

What is Resultant Velocity?

Resultant velocity is the total, actual velocity of an object when two or more independent velocity vectors act on it at the same time. The most classic scenario involves an object moving through a medium that is itself moving—such as an aeroplane flying through moving air (wind), or a swimmer crossing a flowing river.

The fundamental rule of resultant velocity is vector addition:
\(\mathbf{v}_{\text{resultant}} = \mathbf{v}_{\text{engine/still water}} + \mathbf{v}_{\text{wind/current}}\)

Key Terminology in Navigation Problems

When tackling examination questions, you will encounter specific terms that tell you which vectors you are dealing with:

• Velocity in still water / Airspeed: The velocity produced solely by the boat's engine or the aeroplane's propeller. This is the direction the craft is pointing or steered.
• Current / Wind velocity: The velocity of the medium itself.
• Course made good / Actual track / Resultant velocity: The real path and speed of the craft as observed by someone standing stationary on the riverbank or on the ground.

Setting Up Vector Components

In 2D mechanics, we almost always resolve velocities into standard Cartesian unit vectors \(\mathbf{i}\) (directed East) and \(\mathbf{j}\) (directed North):

• A bearing of \(\theta^\circ\) (measured clockwise from North) gives:
\(\mathbf{v} = (v \sin\theta)\mathbf{i} + (v \cos\theta)\mathbf{j}\)
• The magnitude (speed) is \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\)
• The direction angle \(\alpha\) measured from the positive x-axis is given by \(\tan\alpha = \frac{v_y}{v_x}\)

Worked Example: Crossing a Flowing River

A boat sets out from a point on the south bank of a river. The river flows due East at \(3\text{ m s}^{-1}\). The boat's engine gives it a speed of \(5\text{ m s}^{-1}\) in still water. If the boat heads directly North across the river, find its resultant speed and direction.

Step 1: Write down the individual velocity vectors.
Velocity of water: \(\mathbf{v}_w = 3\mathbf{i}\text{ m s}^{-1}\)
Velocity of boat relative to water: \(\mathbf{v}_b = 5\mathbf{j}\text{ m s}^{-1}\)

Step 2: Add the vectors to find the resultant velocity.
\(\mathbf{v}_{\text{resultant}} = \mathbf{v}_b + \mathbf{v}_w = 3\mathbf{i} + 5\mathbf{j}\text{ m s}^{-1}\)

Step 3: Calculate the magnitude and bearing.
Resultant speed \(= |\mathbf{v}_{\text{resultant}}| = \sqrt{3^2 + 5^2} = \sqrt{34} \approx 5.83\text{ m s}^{-1}\)
Direction: Angle with North \(\theta\) satisfies \(\tan\theta = \frac{3}{5} = 0.6 \implies \theta \approx 31.0^\circ\)
So the boat moves with a speed of \(5.83\text{ m s}^{-1}\) on a bearing of \(031.0^\circ\).

Key Takeaway

Resultant velocity is simply the vector sum of all velocities acting on a body. Always break vectors into \(\mathbf{i}\) and \(\mathbf{j}\) components to make adding them straightforward!


2. Relative Velocity

Understanding Relative Velocity

Imagine you are sitting in a car travelling at \(60\text{ mph}\) North, and another car alongside you travels at \(65\text{ mph}\) North. To you, looking out the side window, the other car appears to be creeping ahead at just \(5\text{ mph}\). However, a pedestrian standing on the pavement sees both cars flying past at their full speeds!

Relative velocity is the velocity of one moving object as observed from the frame of reference of another moving object.

The Fundamental Formula

The velocity of object \(A\) relative to object \(B\) is written as \({}_A\mathbf{v}_B\) (or \(\mathbf{v}_{A/B}\)):
\({}_A\mathbf{v}_B = \mathbf{v}_A - \mathbf{v}_B\)

Similarly, the velocity of \(B\) relative to \(A\) is:
\({}_B\mathbf{v}_A = \mathbf{v}_B - \mathbf{v}_A = -({}_A\mathbf{v}_B)\)

Memory Aid: The "Subtract the Observer" Rule

To find the velocity of anything relative to \(B\), you must make \(B\) the observer. In the observer's world, \(B\) thinks they are completely stationary. To bring \(B\) to rest mathematically, you subtract \(B\)'s velocity: Always subtract the observer!

Worked Example: Relative Motion of Two Cars

Car \(A\) travels with velocity \(\mathbf{v}_A = (12\mathbf{i} + 16\mathbf{j})\text{ m s}^{-1}\). Car \(B\) travels with velocity \(\mathbf{v}_B = (-8\mathbf{i} + 6\mathbf{j})\text{ m s}^{-1}\). Find the velocity of Car \(A\) relative to Car \(B\), and its relative speed.

Step 1: Apply the relative velocity formula.
\({}_A\mathbf{v}_B = \mathbf{v}_A - \mathbf{v}_B\)
\({}_A\mathbf{v}_B = (12\mathbf{i} + 16\mathbf{j}) - (-8\mathbf{i} + 6\mathbf{j})\)
\({}_A\mathbf{v}_B = (12 - (-8))\mathbf{i} + (16 - 6)\mathbf{j} = (20\mathbf{i} + 10\mathbf{j})\text{ m s}^{-1}\)

Step 2: Calculate the relative speed.
Relative speed \(= |{}_A\mathbf{v}_B| = \sqrt{20^2 + 10^2} = \sqrt{500} = 10\sqrt{5} \approx 22.4\text{ m s}^{-1}\)

Key Takeaway

To find the velocity of \(A\) relative to \(B\), compute \({}_A\mathbf{v}_B = \mathbf{v}_A - \mathbf{v}_B\). In this frame of reference, \(B\) is stationary at the origin, and \(A\) appears to move along the vector \({}_A\mathbf{v}_B\).


3. Position and Closest Approach

What is Closest Approach?

In radar tracking and maritime navigation, two vessels on intersecting or passing courses need to know if they risk a collision, how close they will get to each other, and at what time this minimum distance occurs.

We can solve closest approach problems using two powerful methods:

Method 1: The Vector / Calculus Method

Let the initial position vectors of \(A\) and \(B\) at time \(t = 0\) be \(\mathbf{r}_{A0}\) and \(\mathbf{r}_{B0}\). If both move with constant velocities \(\mathbf{v}_A\) and \(\mathbf{v}_B\), their positions at any time \(t\) are:
\(\mathbf{r}_A(t) = \mathbf{r}_{A0} + \mathbf{v}_A t\)
\(\mathbf{r}_B(t) = \mathbf{r}_{B0} + \mathbf{v}_B t\)

The relative displacement vector from \(B\) to \(A\) at time \(t\) is:
\(\mathbf{r}(t) = \mathbf{r}_A(t) - \mathbf{r}_B(t) = (\mathbf{r}_{A0} - \mathbf{r}_{B0}) + (\mathbf{v}_A - \mathbf{v}_B)t = \mathbf{r}_0 + {}_A\mathbf{v}_B t\)

The distance squared between them is \(d^2 = |\mathbf{r}(t)|^2 = x(t)^2 + y(t)^2\).
To find the time \(t\) of closest approach, differentiate \(d^2\) with respect to \(t\) and set \(\frac{d(d^2)}{dt} = 0\), then substitute \(t\) back in to find the minimum distance \(d_{\min}\).

Method 2: The Geometric / Relative Path Method

Imagine freezing \(B\) at the origin. \(A\) starts at the relative position \(\mathbf{r}_0 = \mathbf{r}_{A0} - \mathbf{r}_{B0}\) and travels in a straight line along the direction of the relative velocity \({}_A\mathbf{v}_B\).

• The shortest distance from \(B\) to the straight line path of \(A\) is the perpendicular distance from the origin to that line.
• If \(\theta\) is the angle between \(\mathbf{r}_0\) and \({}_A\mathbf{v}_B\), then:
\(d_{\min} = |\mathbf{r}_0| \sin\theta\)
• The distance along the relative path to the closest point is \(|\mathbf{r}_0| \cos\theta\), so the time taken is:
\(t = \frac{|\mathbf{r}_0| \cos\theta}{|{}_A\mathbf{v}_B|}\)

Worked Example: Finding Shortest Distance and Time

At \(t = 0\), Ship \(A\) is at the origin \((0\mathbf{i} + 0\mathbf{j})\text{ km}\) moving with velocity \(\mathbf{v}_A = (8\mathbf{i} + 4\mathbf{j})\text{ km h}^{-1}\). Ship \(B\) is at \((10\mathbf{i} + 0\mathbf{j})\text{ km}\) moving with velocity \(\mathbf{v}_B = (2\mathbf{i} + 12\mathbf{j})\text{ km h}^{-1}\). Find the time of closest approach and the shortest distance between them.

Step 1: Express relative displacement \(\mathbf{r}_{A/B}(t)\).
Initial relative position: \(\mathbf{r}_0 = \mathbf{r}_{A0} - \mathbf{r}_{B0} = (0\mathbf{i} + 0\mathbf{j}) - (10\mathbf{i} + 0\mathbf{j}) = -10\mathbf{i}\text{ km}\)
Relative velocity: \({}_A\mathbf{v}_B = \mathbf{v}_A - \mathbf{v}_B = (8\mathbf{i} + 4\mathbf{j}) - (2\mathbf{i} + 12\mathbf{j}) = (6\mathbf{i} - 8\mathbf{j})\text{ km h}^{-1}\)

At time \(t\), the displacement of \(A\) relative to \(B\) is:
\(\mathbf{r}(t) = -10\mathbf{i} + (6\mathbf{i} - 8\mathbf{j})t = (6t - 10)\mathbf{i} - 8t\mathbf{j}\)

Step 2: Find the distance squared \(d^2\).
\(d^2 = (6t - 10)^2 + (-8t)^2\)
\(d^2 = 36t^2 - 120t + 100 + 64t^2 = 100t^2 - 120t + 100\)

Step 3: Minimise \(d^2\) with respect to \(t\).
\(\frac{d(d^2)}{dt} = 200t - 120 = 0 \implies t = \frac{120}{200} = 0.6\text{ hours}\) (i.e. \(36\text{ minutes}\))

Step 4: Calculate the minimum distance.
Substitute \(t = 0.6\) into \(d^2\):
\(d^2 = 100(0.6)^2 - 120(0.6) + 100 = 100(0.36) - 72 + 100 = 36 - 72 + 100 = 64\)
\(d_{\min} = \sqrt{64} = 8\text{ km}\)

Key Takeaway

Closest approach problems can be solved cleanly by expressing the relative position vector as a function of \(t\), forming an expression for \(d^2\), and differentiating to find the stationary point.


4. Interception and Collision

Condition for Collision / True Interception

Two objects \(A\) and \(B\) will collide (or intercept) if they arrive at the exact same location at the exact same time \(T\):
\(\mathbf{r}_A(T) = \mathbf{r}_B(T)\)

In terms of relative motion, this means the relative position vector at time \(T\) must be zero:
\(\mathbf{r}_0 + {}_A\mathbf{v}_B T = \mathbf{0} \implies {}_A\mathbf{v}_B = -\frac{\mathbf{r}_0}{T}\)

Vital Insight: The relative velocity \({}_A\mathbf{v}_B\) must point directly along the line joining their initial positions from \(A\) to \(B\) (i.e. in the direction of \(\mathbf{r}_B - \mathbf{r}_A\)).

Constant Bearing: The Sailor's Rule

Did you know? In maritime navigation, if you look at an approaching ship and its bearing from you does not change over time, you are on a collision course! Mathematically, constant bearing means the direction of the relative position vector is constant and pointing directly along the relative velocity line.

Worked Example: Setting a Course to Intercept

A patrol boat \(P\) is at the origin. A target vessel \(T\) is at position \((12\mathbf{i} + 16\mathbf{j})\text{ km}\) and moving with constant velocity \(\mathbf{v}_T = (4\mathbf{i} + 2\mathbf{j})\text{ km h}^{-1}\). The patrol boat can travel at a speed of \(10\text{ km h}^{-1}\). Determine the velocity vector of \(P\) required to intercept \(T\), and find the time taken for interception.

Step 1: Set up the position vectors at interception time \(t\).
\(\mathbf{r}_T(t) = (12 + 4t)\mathbf{i} + (16 + 2t)\mathbf{j}\)
Let \(\mathbf{v}_P = u\mathbf{i} + v\mathbf{j}\), where \(u^2 + v^2 = 10^2 = 100\).
\(\mathbf{r}_P(t) = (ut)\mathbf{i} + (vt)\mathbf{j}\)

Step 2: Equate positions at collision time \(t\).
\(ut = 12 + 4t \implies ut - 4t = 12 \implies (u - 4)t = 12\)
\(vt = 16 + 2t \implies vt - 2t = 16 \implies (v - 2)t = 16\)

Step 3: Eliminate \(t\) to find a relation between \(u\) and \(v\).
\(t = \frac{12}{u - 4} = \frac{16}{v - 2} \implies 12(v - 2) = 16(u - 4)\)
Divide by \(4\): \(3(v - 2) = 4(u - 4) \implies 3v - 6 = 4u - 16 \implies 3v = 4u - 10 \implies v = \frac{4u - 10}{3}\)

Step 4: Substitute into the speed equation \(u^2 + v^2 = 100\).
\(u^2 + \left(\frac{4u - 10}{3}\right)^2 = 100\)
\(u^2 + \frac{16u^2 - 80u + 100}{9} = 100\)
Multiply by \(9\): \(9u^2 + 16u^2 - 80u + 100 = 900\)
\(25u^2 - 80u - 800 = 0\)
Divide by \(5\): \(5u^2 - 16u - 160 = 0\)

Using the quadratic formula for \(u\):
\(u = \frac{16 \pm \sqrt{(-16)^2 - 4(5)(-160)}}{2(5)} = \frac{16 \pm \sqrt{256 + 3200}}{10} = \frac{16 \pm \sqrt{3456}}{10} = \frac{16 \pm 58.788}{10}\)

Since the target has a positive x-coordinate, \(u\) must be positive to intercept in positive time:
\(u = \frac{16 + 58.788}{10} \approx 7.48\text{ km h}^{-1}\)
Then \(v = \frac{4(7.479) - 10}{3} \approx 6.64\text{ km h}^{-1}\)

Step 5: Find the time taken \(t\).
\(t = \frac{12}{u - 4} = \frac{12}{7.479 - 4} = \frac{12}{3.479} \approx 3.45\text{ hours}\)

Hence, the required velocity of the patrol boat is \(\mathbf{v}_P \approx (7.48\mathbf{i} + 6.64\mathbf{j})\text{ km h}^{-1}\), intercepting in approximately \(3.45\text{ hours}\).

Key Takeaway

For an interception problem, equate the displacement vectors of both objects at time \(t\). If the speed of the interceptor is fixed, solve the resulting system of equations to determine the unknown velocity components and time.


5. Common Mistakes to Avoid

• Mixing up the Order of Subtraction: Remember that the velocity of \(A\) relative to \(B\) is \(\mathbf{v}_A - \mathbf{v}_B\), NOT \(\mathbf{v}_B - \mathbf{v}_A\). Getting the sign wrong inverts the relative trajectory!

• Confusing Bearings with Cartesian Angles: Bearings are measured clockwise from North (\(\mathbf{j}\)). Standard Cartesian angles are measured anticlockwise from the positive x-axis (\(\mathbf{i}\)). If a bearing is \(\theta\), the vector is \((v \sin\theta)\mathbf{i} + (v \cos\theta)\mathbf{j}\).

• Adding Speeds Directly as Scalars: Never simply add or subtract magnitudes unless the objects are moving in the exact same straight line! Always use vector components (\(\mathbf{i}, \mathbf{j}\)).

• Forgetting Units: Always check if speeds are given in \(\text{km h}^{-1}\) while times are in minutes or distances in metres. Convert all quantities to consistent SI units before calculating.


Quick Review Summary Checklist

• Resultant Velocity: \(\mathbf{v}_{\text{actual}} = \mathbf{v}_{\text{engine}} + \mathbf{v}_{\text{medium}}\)
• Relative Velocity Formula: \({}_A\mathbf{v}_B = \mathbf{v}_A - \mathbf{v}_B\)
• Relative Position at time \(t\): \(\mathbf{r}(t) = (\mathbf{r}_{A0} - \mathbf{r}_{B0}) + ({}_A\mathbf{v}_B)t\)
• Closest Approach: Form \(d^2(t) = |\mathbf{r}(t)|^2\), differentiate and set \(\frac{d(d^2)}{dt} = 0\)
• Interception: Solve \(\mathbf{r}_A(t) = \mathbf{r}_B(t)\) simultaneously for \(t > 0\)