How to Calculate One Boat’s Velocity Relative to Another

Two boats can be carried downstream by the same river current, yet a passenger in one boat might see the other moving sideways or approaching diagonally. How can that happen when both boats are travelling through the same water?

The answer depends on the observer. A person standing on the riverbank measures each boat’s velocity relative to the ground. A passenger sitting in boat B measures boat A’s velocity relative to B.

To calculate the velocity of boat A as seen from boat B, subtract B’s velocity vector from A’s velocity vector:

\[ \boxed{\vec v_{A/B}=\vec v_{A/G}-\vec v_{B/G}} \]

Here, both velocities on the right-hand side must be measured relative to the same reference frame, G (the ground).

The result is a relative velocity vector. Its magnitude is the relative speed: how quickly the distance between the boats is changing in the direction of their relative motion.

There is an especially useful shortcut for boats in a uniform river current: if both boats experience exactly the same current, that current cancels when calculating their relative velocity.

Let’s work out why, and then use it to solve some less obvious HSC-style problems.

01Identify the observer and the reference frame

Before substituting numbers, establish exactly whose motion you are describing.

Consider three possible reference frames:

The notation tells you which object is moving and which object is acting as the observer.

For example:

• \(\vec v_{A/G}\): velocity of boat A relative to the ground.
• \(\vec v_{A/W}\): velocity of boat A relative to the water.
• \(\vec v_{W/G}\): velocity of the water relative to the ground.
• \(\vec v_{A/B}\): velocity of boat A relative to boat B.

The order matters. The first letter identifies the moving object, and the second identifies the observer.

Therefore, A relative to B is not the same as B relative to A:

\[ \vec v_{B/A}=-\vec v_{A/B} \]

They have equal relative speeds but opposite relative velocity directions.

The relationship between boat, water, and ground

A boat moving through a river has two contributions to its ground velocity:

1. Its velocity relative to the water, produced by its motion through the water.
2. The velocity of the water relative to the ground, produced by the river current.

These combine as vectors:

\[ \boxed{\vec v_{A/G}=\vec v_{A/W}+\vec v_{W/G}} \]

This is the same idea used when calculating the ground velocity of a boat crossing a river.

However, our problem has an extra layer: we want to know how one moving boat sees another moving boat.

02Why the river current can cancel

Imagine two people standing on the same moving walkway.

The walkway carries both people forwards. However, one person walks forwards along the walkway while the other walks sideways relative to it.

Someone watching from the stationary floor sees both people being carried forwards. Someone standing on the walkway sees only the difference between their walking velocities.

A river current works in the same way.

The water acts like the moving walkway. Both boats receive the same additional velocity from the current, provided they experience the same uniform current.

Let’s derive this rather than simply memorising it.

For boat A:

\[ \vec v_{A/G}=\vec v_{A/W}+\vec v_{W/G} \]

For boat B:

\[ \vec v_{B/G}=\vec v_{B/W}+\vec v_{W/G} \]

We want A’s velocity relative to B:

\[ \vec v_{A/B}=\vec v_{A/G}-\vec v_{B/G} \]

Substituting the two ground velocities:

\[ \begin{aligned} \vec v_{A/B} &=(\vec v_{A/W}+\vec v_{W/G})\\ &\quad-(\vec v_{B/W}+\vec v_{W/G}) \end{aligned} \]

The shared current appears once with a positive sign and once with a negative sign.

It therefore cancels:

\[ \boxed{\vec v_{A/B}=\vec v_{A/W}-\vec v_{B/W}} \]

This is a very useful shortcut.

If both boats are moving in the same uniform current, you can calculate their relative velocity directly from their velocities relative to the water.

There is one important limitation. The shortcut only works when the current velocity is the same at both boats at the same time. If the current varies across the river, the current contributions might not cancel.

The moving-walkway analogy also has a limit: it assumes the walkway moves uniformly. A river with changing current speed is more complicated.

03An experimental way to understand relative boat velocity

How could we test whether a uniform current actually changes the motion of one boat as seen from another?

Consider a proposed classroom investigation using two small motorised model boats. The following apparatus and measurements are illustrative, rather than a report of a particular historical experiment.

The original measurement problem

Imagine a long, transparent water channel with a pump at one end producing a steady eastward current.

Place two model boats in the channel:

• Boat A moves north relative to the water.
• Boat B moves east relative to the water.

A simple first design might attach speed sensors to each boat.

However, there is a problem. Speed-only measurements tell us how fast each boat moves, but not its direction. They also need to specify whether the measurements are relative to the water or the stationary channel.

We cannot reliably calculate a relative velocity vector from those readings alone.

Revising the apparatus

A better design uses an overhead camera fixed above the channel.

Place a calibrated coordinate grid underneath the water channel. The camera tracks the two boats’ positions at regular time intervals.

A small floating marker is also tracked to measure the water’s movement relative to the stationary channel.

The arrangement now provides three sets of measurements:

• The ground-frame displacement of boat A.
• The ground-frame displacement of boat B.
• The displacement of the water itself.

What should the experiment show?

Suppose the water moves east at a constant speed.

Both boats should receive the same additional eastward velocity. Therefore, their relative velocity should be unchanged compared with a still-water trial, assuming their velocities relative to the water remain constant.

Let’s use some illustrative measurements to check this prediction.

Over a \(0.50\text{ s}\) interval, the camera records:

Calculate the ground-frame velocities using:

\[ \vec v=\frac{\Delta\vec r}{\Delta t} \]

The measured water velocity is:

\[ \vec v_{W/G}= \left(\frac{0.050}{0.50},0\right) \]

\[ \vec v_{W/G}=(0.10,0)\text{ m s}^{-1} \]

Using east as the positive \(x\)-direction and north as the positive \(y\)-direction:

\[ \vec v_{A/G}=(0.10,0.20)\text{ m s}^{-1} \]

\[ \vec v_{B/G}=(0.25,0)\text{ m s}^{-1} \]

The velocity of A relative to B is:

\[ \begin{aligned} \vec v_{A/B} &=(0.10,0.20)-(0.25,0)\\ &=(-0.15,0.20)\text{ m s}^{-1} \end{aligned} \]

Its magnitude is:

\[ \begin{aligned} |\vec v_{A/B}| &=\sqrt{(-0.15)^2+(0.20)^2}\\ &=\sqrt{0.0625}\\ &=0.25\text{ m s}^{-1} \end{aligned} \]

The relative speed is therefore \(0.25\text{ m s}^{-1}\).

Notice that the water’s eastward velocity has disappeared from the final calculation.

The same result follows if we calculate the boats’ velocities relative to the water:

\[ \vec v_{A/W}=(0,0.20) \]

\[ \vec v_{B/W}=(0.15,0) \]

Subtracting:

\[ \vec v_{A/B}=(-0.15,0.20) \]

Both methods agree.

What does the experiment establish?

The revised measurement system supports the prediction: a uniform current adds the same velocity to both boats, leaving their relative velocity unchanged.

There are experimental limitations. The camera measures average velocities over the \(0.50\text{ s}\) interval, rather than exact instantaneous velocities. Also, the floating marker must represent the current experienced by both boats.

If the boats accelerate, turn, or experience different currents, their relative velocity can change.

04Relative velocity is a vector, not a speed difference

This is where many students lose marks.

Speed is a scalar. It tells us how fast something moves, without specifying its direction.

Velocity is a vector. It includes both magnitude and direction.

When calculating relative velocity, we must subtract the velocity vectors, not simply subtract their speed magnitudes.

Choose east as positive \(x\) and north as positive \(y\).

A velocity vector can then be written as:

\[ \vec v=(v_x,v_y) \]

The relative velocity components are:

\[ v_{A/B,x}=v_{A,x}-v_{B,x} \]

\[ v_{A/B,y}=v_{A,y}-v_{B,y} \]

Once we have these components, the relative speed is:

\[ \boxed{ s_{A/B}=\sqrt{v_{A/B,x}^2+v_{A/B,y}^2} } \]

The speed is always non-negative. Negative components simply indicate motion in the negative coordinate directions.

Three situations worth comparing

Assume all velocities are measured in the same reference frame.

In the first two cases, the boats move along the same straight line. Subtracting their signed velocities gives the correct answer.

In the third case, the boats move at right angles.

Subtracting their speed magnitudes would give \(4-3=1\text{ m s}^{-1}\), which is incorrect.

The actual relative speed is \(5\text{ m s}^{-1}\), found using Pythagoras.

07When does the current not cancel?

So far, we have assumed that both boats experience the same current.

But real rivers do not always have uniform flow. Water can move faster in one region than another.

Imagine two boats travelling north through different lanes of a river.

Boat A experiences a northward current of \(2.4\text{ m s}^{-1}\), while boat B experiences a northward current of \(0.9\text{ m s}^{-1}\).

Both boats travel north at \(3.0\text{ m s}^{-1}\) relative to their local water.

Their ground velocities are:

\[ \vec v_{A/G}=(0,5.4) \]

\[ \vec v_{B/G}=(0,3.9) \]

Therefore:

\[ \vec v_{A/B}=(0,5.4)-(0,3.9) \]

\[ \boxed{\vec v_{A/B}=(0,1.5)\text{ m s}^{-1}} \]

Boat A moves \(1.5\text{ m s}^{-1}\) north relative to boat B.

Why didn’t the current cancel?

Because the current velocities were different.

In general, when the boats experience different local currents:

\[ \begin{aligned} \vec v_{A/B} ={}&(\vec v_{A/W_A}+\vec v_{W_A/G})\\ &-(\vec v_{B/W_B}+\vec v_{W_B/G}) \end{aligned} \]

The current terms only cancel if:

\[ \vec v_{W_A/G}=\vec v_{W_B/G} \]

This is an important assumption to check before using the shortcut.

08A reliable method for HSC boat-relative-velocity questions

When you see a question asking for the velocity of one boat as seen from another, work through these decisions in order.

1. Identify the observer. If the question asks for A as seen from B, calculate \(\vec v_{A/B}\), not \(\vec v_{B/A}\).

2. Identify the reference frames. Check whether each velocity is relative to the water or the ground.

3. Choose your axes. For river problems, east-west and north-south components are usually convenient. State which directions are positive.

4. Convert to a common frame. Add the current to each boat’s water-frame velocity if using the ground frame. Alternatively, use the water frame directly if both boats experience the same uniform current.

5. Subtract vectors in the correct order. Always calculate A minus B for A relative to B.

6. Calculate the magnitude and direction. Use Pythagoras for the relative speed and trigonometry for its direction.

7. Check the physical meaning. Does the direction make sense from the observer’s point of view? Is the speed positive? Have you mixed reference frames?

If you need to refresh the one-dimensional subtraction method, read Relative Velocity in One Dimension for HSC Physics.

For the different problem of finding a boat’s ground track, direction, and crossing time in a current, see Boats Crossing a Current: Velocity, Track and Crossing Time.

09Questions and solutions

The following questions build from straightforward relative motion to situations where you need to think carefully about reference frames, current variation, and the actual direction of motion.

Question 1

A river current flows east at \(1.2\text{ m s}^{-1}\).

Boat A travels north at \(5.0\text{ m s}^{-1}\) relative to the water. Boat B travels north at \(3.0\text{ m s}^{-1}\) relative to the water.

Both boats experience the same uniform current and maintain constant velocities.

Calculate:

(a) The velocity of boat A relative to boat B.

(b) The relative speed of boat A as measured by a passenger in boat B.

Solution 1

(a) Boat A moves north at \(2.0\text{ m s}^{-1}\) relative to boat B. (b) The relative speed is \(2.0\text{ m s}^{-1}\).

Both boats experience the same eastward current, so it cancels.

Their ground velocities are:

\[ \vec v_{A/G}=(1.2,5.0) \]

\[ \vec v_{B/G}=(1.2,3.0) \]

Subtract:

\[ \begin{aligned} \vec v_{A/B} &=(1.2,5.0)-(1.2,3.0)\\ &=(0,2.0)\text{ m s}^{-1} \end{aligned} \]

The zero eastward component means there is no relative east-west motion.

The positive northward component means boat A moves north relative to B.

Therefore:

\[ \boxed{\vec v_{A/B}=2.0\text{ m s}^{-1}\text{ north}} \]

Because the relative velocity is entirely northward, its magnitude is also \(2.0\text{ m s}^{-1}\).

The important observation is that the current affects both ground velocities but does not affect their difference.

Question 2

A uniform river current flows east at \(2.0\text{ m s}^{-1}\).

Two boats move through the water at the same speed of \(5.0\text{ m s}^{-1}\).

• Boat A travels at \(53.1^\circ\) north of east relative to the water.
• Boat B travels at \(36.9^\circ\) south of east relative to the water.

Calculate the velocity of boat A relative to boat B, including its magnitude and direction.

Explain why the relative speed is not zero, even though both boats have the same speed relative to the water.

Solution 2

Boat A moves relative to boat B at \(7.07\text{ m s}^{-1}\), directed \(8.13^\circ\) west of north.

The boats have equal speeds, but their velocity directions are different. Equal speed does not mean equal velocity.

Since both experience the same uniform current, we can calculate their relative velocity directly using their water-frame velocities.

Step 1: Resolve boat A’s velocity into components.

\[ \begin{aligned} v_{A/W,x}&=5.0\cos53.1^\circ=3.0\\ v_{A/W,y}&=5.0\sin53.1^\circ=4.0 \end{aligned} \]

Therefore:

\[ \vec v_{A/W}=(3.0,4.0) \]

Step 2: Resolve boat B’s velocity into components.

\[ \begin{aligned} v_{B/W,x}&=5.0\cos36.9^\circ=4.0\\ v_{B/W,y}&=-5.0\sin36.9^\circ=-3.0 \end{aligned} \]

Therefore:

\[ \vec v_{B/W}=(4.0,-3.0) \]

The negative north-south component indicates southward motion.

Step 3: Subtract the vectors.

\[ \begin{aligned} \vec v_{A/B} &=(3.0,4.0)-(4.0,-3.0)\\ &=(-1.0,7.0)\text{ m s}^{-1} \end{aligned} \]

Step 4: Calculate the relative speed.

\[ \begin{aligned} s_{A/B} &=\sqrt{(-1.0)^2+7.0^2}\\ &=\sqrt{50}\\ &=7.07\text{ m s}^{-1} \end{aligned} \]

Step 5: Calculate the direction.

The vector points north-west, with a westward component of \(1.0\text{ m s}^{-1}\) and a northward component of \(7.0\text{ m s}^{-1}\).

Measure the angle west of north:

\[ \theta=\tan^{-1}\left(\frac{1.0}{7.0}\right) \]

\[ \theta=8.13^\circ \]

Therefore:

\[ \boxed{ \vec v_{A/B}=7.07\text{ m s}^{-1} \text{ at }8.13^\circ \text{ west of north} } \]

The tempting mistake is to subtract the two speed magnitudes and obtain zero.

That would only be valid if the velocity vectors were identical. Here, the boats travel in different directions, so their relative velocity is substantial.

Question 3

A shore-based GPS measures boat A travelling at \(5.0\text{ m s}^{-1}\), \(36.9^\circ\) north of east.

A passenger travelling in boat B measures boat A’s velocity as:

• \(7.0\text{ m s}^{-1}\) east,
• \(3.0\text{ m s}^{-1}\) north.

The river current flows east at \(1.0\text{ m s}^{-1}\).

Assuming constant velocities and a uniform current, determine the velocity of boat B relative to the water.

Solution 3

Boat B travels west at \(4.0\text{ m s}^{-1}\) relative to the water.

This is an inverse relative-velocity problem. Instead of calculating the relative velocity from two known boat velocities, we are given the relative velocity and need to work backwards.

Step 1: Calculate boat A’s ground-frame velocity.

Resolve its velocity into components:

\[ \begin{aligned} v_{A/G,x}&=5.0\cos36.9^\circ=4.0\\ v_{A/G,y}&=5.0\sin36.9^\circ=3.0 \end{aligned} \]

Therefore:

\[ \vec v_{A/G}=(4.0,3.0) \]

Step 2: Write the observed relative velocity.

The passenger in B measures A moving east at \(7.0\text{ m s}^{-1}\) and north at \(3.0\text{ m s}^{-1}\).

Therefore:

\[ \vec v_{A/B}=(7.0,3.0) \]

Step 3: Rearrange the relative-velocity equation.

We know:

\[ \vec v_{A/B}=\vec v_{A/G}-\vec v_{B/G} \]

Rearranging:

\[ \vec v_{B/G}=\vec v_{A/G}-\vec v_{A/B} \]

Substitute:

\[ \begin{aligned} \vec v_{B/G} &=(4.0,3.0)-(7.0,3.0)\\ &=(-3.0,0) \end{aligned} \]

So boat B travels west at \(3.0\text{ m s}^{-1}\) relative to the ground.

Step 4: Convert to the water frame.

The relationship is:

\[ \vec v_{B/G}=\vec v_{B/W}+\vec v_{W/G} \]

Rearrange:

\[ \vec v_{B/W}=\vec v_{B/G}-\vec v_{W/G} \]

Substitute:

\[ \begin{aligned} \vec v_{B/W} &=(-3.0,0)-(1.0,0)\\ &=(-4.0,0)\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{ \vec v_{B/W}=4.0\text{ m s}^{-1}\text{ west} } \]

The important distinction is that boat B moves west at \(3.0\text{ m s}^{-1}\) relative to the ground but must move west at \(4.0\text{ m s}^{-1}\) relative to the water.

The eastward current reduces its westward ground speed.

Question 4

A river flows north, but its current is not uniform across its width.

Two boats travel in separate parallel lanes:

• Boat A experiences a northward current of \(2.4\text{ m s}^{-1}\).
• Boat B experiences a northward current of \(0.9\text{ m s}^{-1}\).

Both boats travel north at \(3.0\text{ m s}^{-1}\) relative to their local water.

At \(t=0\), the boats are abreast, meaning they have the same north-south position. They remain in their respective lanes and maintain constant velocities for the next \(12\text{ s}\).

Calculate:

(a) The velocity of boat A relative to boat B.

(b) How far ahead of boat B boat A will be after \(12\text{ s}\), measured along the river.

(c) Explain why the river current does not cancel in this situation.

Solution 4

(a) Boat A moves north at \(1.5\text{ m s}^{-1}\) relative to boat B. (b) After \(12\text{ s}\), boat A is \(18\text{ m}\) farther downstream. (c) The current does not cancel because the boats experience different current velocities.

This question tests the assumption behind the shared-current shortcut.

Step 1: Calculate boat A’s ground velocity.

Its through-water velocity and local current are both northward:

\[ \begin{aligned} v_{A/G}&=3.0+2.4\\ &=5.4\text{ m s}^{-1}\text{ north} \end{aligned} \]

Step 2: Calculate boat B’s ground velocity.

\[ \begin{aligned} v_{B/G}&=3.0+0.9\\ &=3.9\text{ m s}^{-1}\text{ north} \end{aligned} \]

Step 3: Calculate the relative velocity.

Both boats travel north, so we can subtract their signed velocities:

\[ \begin{aligned} v_{A/B}&=5.4-3.9\\ &=1.5\text{ m s}^{-1}\text{ north} \end{aligned} \]

Therefore:

\[ \boxed{\vec v_{A/B}=1.5\text{ m s}^{-1}\text{ north}} \]

Step 4: Calculate the downstream separation after \(12\text{ s}\).

The boats initially have the same north-south position.

Their relative downstream displacement is:

\[ \Delta y=v_{A/B}t \]

Substitute:

\[ \begin{aligned} \Delta y&=1.5(12)\\ &=18\text{ m} \end{aligned} \]

Therefore, boat A is \(18\text{ m}\) farther downstream than boat B.

This is their downstream separation change, not their total distance apart, because they remain in separate lanes.

Step 5: Explain why the current does not cancel.

The tempting assumption is that both boats are travelling through water, so the current should disappear.

However, the current at A is \(2.4\text{ m s}^{-1}\), while the current at B is only \(0.9\text{ m s}^{-1}\).

The difference is:

\[ 2.4-0.9=1.5\text{ m s}^{-1} \]

That difference contributes directly to their relative velocity.

The shared-current shortcut cannot be used when the local current vectors are different.

Question 5

Two boats travel on a wide river with a uniform eastward current of \(1.0\text{ m s}^{-1}\).

At \(t=0\):

• Boat A is \(120\text{ m}\) directly north of boat B.
• Boat A travels east at \(3.0\text{ m s}^{-1}\) relative to the water.
• Boat B travels north at \(4.0\text{ m s}^{-1}\) relative to the water.

Both boats maintain constant velocities, and the current is uniform throughout the region.

A student calculates the relative speed of the boats as \(5.0\text{ m s}^{-1}\), then divides their initial separation by this speed:

\[ t=\frac{120}{5.0}=24\text{ s} \]

The student concludes that the boats will collide after \(24\text{ s}\).

Determine whether this conclusion is correct.

If the boats do not collide, calculate their minimum separation and the time at which this occurs.

Solution 5

The boats do not collide. Their minimum separation is \(72\text{ m}\), occurring \(19.2\text{ s}\) after the initial observation.

The student’s relative speed calculation is correct, but the collision prediction is not.

The mistake is assuming that the relative velocity points directly along the initial separation between the boats.

Relative speed tells us how quickly the relative position vector changes in magnitude and direction. It does not tell us whether the boats are moving directly towards one another.

We need to consider both components of their relative motion.

Step 1: Calculate the ground velocities.

Choose east as positive \(x\) and north as positive \(y\).

For boat A:

\[ \begin{aligned} \vec v_{A/G} &=(3.0,0)+(1.0,0)\\ &=(4.0,0)\text{ m s}^{-1} \end{aligned} \]

For boat B:

\[ \begin{aligned} \vec v_{B/G} &=(0,4.0)+(1.0,0)\\ &=(1.0,4.0)\text{ m s}^{-1} \end{aligned} \]

Step 2: Calculate the relative velocity.

\[ \begin{aligned} \vec v_{A/B} &=(4.0,0)-(1.0,4.0)\\ &=(3.0,-4.0)\text{ m s}^{-1} \end{aligned} \]

Therefore, boat A moves east at \(3.0\text{ m s}^{-1}\) and south at \(4.0\text{ m s}^{-1}\) relative to boat B.

Its relative speed is:

\[ \begin{aligned} s_{A/B} &=\sqrt{3.0^2+(-4.0)^2}\\ &=5.0\text{ m s}^{-1} \end{aligned} \]

The relative speed is indeed \(5.0\text{ m s}^{-1}\).

Step 3: Write the initial relative position.

At \(t=0\), boat A is \(120\text{ m}\) north of boat B.

Taking boat B as the origin, the initial relative position of A is:

\[ \vec r_{A/B}(0)=(0,120) \]

After \(t\) seconds, the relative position becomes:

\[ \vec r_{A/B}(t)=\vec r_{A/B}(0)+\vec v_{A/B}t \]

Substituting:

\[ \boxed{ \vec r_{A/B}(t)=(3t,120-4t) } \]

The first component describes their east-west separation. The second describes their north-south separation.

Step 4: Determine whether a collision is possible.

For a collision, both components of their relative position must equal zero at the same time.

The east-west component requires:

\[ 3t=0 \]

Therefore:

\[ t=0 \]

The north-south component requires:

\[ 120-4t=0 \]

Therefore:

\[ t=30\text{ s} \]

These conditions cannot both be satisfied at the same time.

The boats do not collide.

At \(t=30\text{ s}\), they have the same north-south position, but their east-west separation is \(90\text{ m}\).

Step 5: Calculate the minimum separation.

The distance between the boats at time \(t\) is:

\[ d=\sqrt{(3t)^2+(120-4t)^2} \]

Rather than repeatedly calculating distances, square both sides:

\[ d^2=9t^2+(120-4t)^2 \]

Expand:

\[ d^2=9t^2+14400-960t+16t^2 \]

Collect like terms:

\[ d^2=25t^2-960t+14400 \]

Complete the square:

\[ \boxed{ d^2=25(t-19.2)^2+5184 } \]

The squared term cannot be negative. Its smallest possible value is zero, which occurs when:

\[ t-19.2=0 \]

Therefore:

\[ t=19.2\text{ s} \]

At this time:

\[ d^2_{\min}=5184 \]

Taking the square root:

\[ d_{\min}=\sqrt{5184} \]

\[ \boxed{d_{\min}=72\text{ m}} \]

The student’s tempting calculation,

\[ \frac{120}{5.0}=24\text{ s} \]

assumes the boats close directly along their initial north-south separation.

But boat A also moves east relative to boat B. This sideways motion prevents the boats from meeting.

The key distinction is between relative speed and the direction of relative velocity.

A boat can have a large relative speed without moving directly towards another boat.

This is why collision and interception problems require both the relative velocity vector and the initial separation vector, not just their magnitudes.

Once you can calculate relative velocity confidently, the next step is using it to predict whether two moving objects will meet, pass one another, or maintain a constant separation.