How to Draw a Relative-Velocity Triangle Without Reversing an Arrow
The vector triangle isn't just a drawing convention. It describes something that can be measured.
Two carts move across a laboratory floor. Cart A travels north, while cart B travels east. A camera attached to cart B records A moving diagonally.
You draw both carts' velocity arrows from the same starting point. Now you need to draw the velocity of A relative to B. Which way should the final arrow point?
Here's the rule that prevents the most common mistake:
**Draw the relative-velocity arrow from the tip of the observer's velocity vector to the tip of the object's velocity vector.**
Not the other way around.
This gives you the correct direction without needing to reverse either of the original arrows. Once you understand why, drawing relative-velocity triangles becomes much more reliable.
[IMAGE: Overhead illustration of two small motorised carts on a laboratory floor with perpendicular grid directions. Cart A travels north and cart B travels east. Show a stationary ground reference, the two carts, and their separate ground-frame velocity arrows. Do not show the relative-velocity arrow yet.]
## 1. Identify the object, the observer, and the reference frame
Before drawing anything, identify three things:
- The **object** is what you want to describe the motion of.
- The **observer** is the object from whose viewpoint you are measuring that motion.
- The **reference frame** is the coordinate system in which the original velocities are measured.
We'll use three labels:
| Symbol | Meaning |
|---|---|
| A | Object whose motion we're investigating |
| B | Observer measuring A's motion |
| G | Ground frame, stationary relative to the laboratory |
For example:
- A is a drone travelling north.
- B is another drone travelling east.
- G is an observer standing on the ground.
The velocity of A relative to B is written as
\[
\vec v_{A/B}
\]
Read this as "the velocity of A as measured by B".
The order matters. A is the object, and B is the observer.
The ground-frame velocities are written as:
- \(\vec v_{A/G}\): velocity of A relative to the ground.
- \(\vec v_{B/G}\): velocity of B relative to the ground.
Both original velocities must be measured in the same reference frame before you can compare them.
## 2. The equation behind the triangle
The relative velocity of A as measured by B is
\[
\boxed{\vec v_{A/B}
=
\vec v_{A/G}-\vec v_{B/G}}
\]
In words:
**Object's ground velocity minus observer's ground velocity.**
This is vector subtraction, not ordinary subtraction of speeds.
You can rearrange the equation:
\[
\boxed{\vec v_{A/G}
=
\vec v_{B/G}+\vec v_{A/B}}
\]
This second version is particularly useful when drawing a head-to-tail velocity triangle.
It says that the object's ground velocity is the vector sum of:
1. The observer's ground velocity.
2. The object's velocity relative to the observer.
### The correct way to draw the arrows
There are two reliable constructions. They represent exactly the same relationship.
**Method 1: Draw both ground velocities from the same starting point.**
1. Draw \(\vec v_{A/G}\) from an arbitrary starting point.
2. Draw \(\vec v_{B/G}\) from that same starting point.
3. Join the arrowheads.
4. Draw the relative-velocity arrow from the tip of \(\vec v_{B/G}\) towards the tip of \(\vec v_{A/G}\).
The direction of that final arrow is important.
It must point from the observer's arrowhead to the object's arrowhead.
[IMAGE: Annotated vector diagram with a common origin O. Draw the ground velocity of observer B horizontally east to point B, and the ground velocity of object A vertically north to point A. Draw the relative velocity arrow from point B to point A, pointing northwest. Label all three vectors clearly, with the relative vector labelled v_A/B. Show the right-angle geometry and make the arrowheads unmistakable.]
**Method 2: Build a head-to-tail triangle.**
Use the rearranged equation:
\[
\vec v_{A/G}
=
\vec v_{B/G}+\vec v_{A/B}
\]
1. Draw \(\vec v_{B/G}\) first.
2. Start \(\vec v_{A/B}\) at the arrowhead of \(\vec v_{B/G}\).
3. Draw the resultant \(\vec v_{A/G}\) from the original starting point to the final arrowhead.
The three vectors now form a triangle.
Notice that the observer's velocity arrow has not been reversed. The relative-velocity arrow is simply the second side of the triangle.
[IMAGE: Clean head-to-tail relative-velocity triangle. Show the observer's ground velocity v_B/G pointing east from O to B, the relative velocity v_A/B pointing northwest from B to A, and the resultant object ground velocity v_A/G pointing north from O to A. Include clear arrowheads and labels. Preserve the correct vector directions.]
### Why students sometimes reverse the arrow
The original equation contains a minus sign:
\[
\vec v_{A/B}=\vec v_{A/G}-\vec v_{B/G}
\]
A common approach is to turn the subtraction into addition:
\[
\vec v_{A/B}
=
\vec v_{A/G}+(-\vec v_{B/G})
\]
This is mathematically correct. However, it means you must reverse the observer's velocity vector before adding it.
That is a valid alternative construction, but it creates an opportunity for mistakes.
For example, a student might reverse the observer's arrow to account for the minus sign, then reverse it again when arranging the triangle. The final relative velocity points in exactly the wrong direction.
The safer approach for most HSC questions is to draw both original ground velocities from a common starting point, then connect the arrowheads in the correct order.
**Observer's arrowhead first. Object's arrowhead second.**
## 3. Velocity is a vector; speed is a scalar
This distinction is essential when drawing relative-velocity triangles.
Speed tells you how fast something moves. It has magnitude but no direction.
Velocity tells you how fast something moves and in which direction. It is a vector.
You can subtract two velocities using vector subtraction. You cannot construct a correct vector triangle using only two speed values.
Consider these two situations:
| Object A | Observer B | Relative velocity of A to B |
|---|---|---|
| 10 m/s east | 6 m/s east | 4 m/s east |
| 10 m/s east | 6 m/s west | 16 m/s east |
The speeds are identical in both situations. The relative velocities are not.
This is why you must establish the directions before doing any calculations.
The magnitude of the relative velocity is the relative speed:
\[
s_{A/B}=|\vec v_{A/B}|
\]
Relative speed cannot be negative. A negative component or signed velocity simply indicates a direction opposite to the chosen positive direction.
If you need a refresher on signed velocities, see [Relative Velocity in One Dimension for HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-relative-velocity-in-one-dimension/).
04An experiment: measuring relative velocity with two moving carts
The vector triangle isn’t just a drawing convention. It describes something that can be measured.
Imagine an investigation designed to answer this question:
Can we predict what an observer travelling with cart B measures for cart A’s velocity?
The original measurement problem
Place two motorised carts on perpendicular tracks across a level laboratory floor.
- Cart A travels north.
- Cart B travels east.
- A sensor on each cart measures its speed.
Suppose the sensors report 8.0 m/s for A and 6.0 m/s for B.
Can we determine the velocity of A relative to B?
Not yet.
The sensors give us two speeds, but they don’t necessarily record direction. The same readings could describe carts travelling in completely different directions.
The original design cannot establish the relative-velocity vector.
Revising the apparatus
Add a marked coordinate grid to the floor and position a fixed overhead camera above the tracks.
Attach a visible marker to each cart. The camera records the coordinates of both markers at regular time intervals.
Choose east as the positive \(x\)-direction and north as the positive \(y\)-direction.
The revised apparatus records the following illustrative data:
| Time (s) | A: x (m) | A: y (m) | B: x (m) | B: y (m) |
|---|---|---|---|---|
| 0.00 | 2.0 | 1.0 | 5.0 | 4.0 |
| 0.50 | 2.0 | 5.0 | 8.0 | 4.0 |
We can now calculate each cart’s velocity.
Step 1: Calculate cart A’s ground-frame velocity.
Over 0.50 s, A moves:
- 0 m east.
- 4.0 m north.
Therefore,
\[ \vec v_{A/G} = \left(0,\frac{4.0}{0.50}\right) \]
\[ \vec v_{A/G} = (0,8.0)\\,\mathrm{m\\,s^{-1}} \]
Step 2: Calculate cart B’s ground-frame velocity.
Over the same interval, B moves:
- 3.0 m east.
- 0 m north.
Therefore,
\[ \vec v_{B/G} = \left(\frac{3.0}{0.50},0\right) \]
\[ \vec v_{B/G} = (6.0,0)\\,\mathrm{m\\,s^{-1}} \]
Step 3: Calculate A’s velocity relative to B.
Subtract B’s velocity from A’s:
\[ \begin{aligned} \vec v_{A/B} &=\vec v_{A/G}-\vec v_{B/G}\\\ &=(0,8.0)-(6.0,0)\\\ &=(-6.0,8.0)\\,\mathrm{m\\,s^{-1}} \end{aligned} \]
The negative \(x\)-component means west. The positive \(y\)-component means north.
So A moves northwest relative to B.
Step 4: Check the result using relative positions.
The position of A relative to B is the difference between their ground-frame positions.
At the beginning:
\[ \vec r_{A/B} = (-3,-3)\\,\mathrm m \]
After 0.50 s:
\[ \vec r_{A/B} = (-6,1)\\,\mathrm m \]
The relative displacement is therefore
\[ \Delta\vec r_{A/B}=(-3,4)\\,\mathrm m \]
Dividing by the elapsed time:
\[ \vec v_{A/B} = \frac{(-3,4)}{0.50} = (-6,8)\\,\mathrm{m\\,s^{-1}} \]
The two methods agree.
The relative velocity arrow must point northwest, from B’s ground-velocity arrowhead towards A’s ground-velocity arrowhead.
The relative speed is
\[ s_{A/B}=\sqrt{6.0^2+8.0^2} =10\\,\mathrm{m\\,s^{-1}} \]
This is an illustrative measurement investigation, not a report of a particular historical experiment.
There is one experimental limitation worth recognising. The position measurements give average velocities over the 0.50 s interval. If either cart accelerates, these averages might not equal its instantaneous velocity at a particular moment. Shorter measurement intervals would help identify changes in velocity.
05A useful decision process for HSC questions
When you encounter a relative-velocity triangle, work through these decisions in order.
| Decision | What to do |
|---|---|
| 1. Identify the object | Decide which object’s velocity you need. |
| 2. Identify the observer | Decide whose viewpoint matters. |
| 3. Identify the common frame | Check that the given velocities use the same reference frame. |
| 4. Write the equation | Object relative to observer = object velocity minus observer velocity. |
| 5. Choose axes | Usually east is positive \(x\), and north is positive \(y\). |
| 6. Draw the vectors | Put both original ground-frame vectors at the same starting point. |
| 7. Draw the relative vector | Start at the observer’s arrowhead and finish at the object’s arrowhead. |
| 8. Calculate | Resolve components, then find magnitude and direction. |
A useful way to check your diagram is to swap the observer and object.
For example,
\[ \vec v_{B/A}=-\vec v_{A/B} \]
The relative-velocity vectors must have equal magnitudes and opposite directions.
If they don’t, revisit your arrow directions.
A contrasting case: when the observer is faster
Suppose A travels east at 10 m/s, while B travels east at 14 m/s.
Both ground-frame arrows point east. However, B’s arrow is longer.
The relative velocity is
\[ \begin{aligned} \vec v_{A/B} &=10-14\\\ &=-4\\,\mathrm{m\\,s^{-1}} \end{aligned} \]
Therefore, A moves at 4 m/s west relative to B.
This is where a common misconception appears.
A student might conclude that A must be travelling west because its relative velocity is negative.
Not so.
A still travels east relative to the ground. It simply moves backwards from B’s viewpoint because B is travelling east faster.
The relative-velocity arrow describes the change in separation between the objects, not necessarily the direction either object travels in the ground frame.
06Worked examples
Worked example: A drone viewed from another moving drone
Drone A travels at 10.0 m/s, at 37° north of east, relative to the ground. Drone B travels east at 4.0 m/s relative to the ground.
Calculate the velocity of A relative to B.
Step 1: Identify the object, observer, and reference frame.
- Object: A.
- Observer: B.
- Reference frame: ground, G.
We need
\[ \vec v_{A/B} = \vec v_{A/G}-\vec v_{B/G} \]
Step 2: Resolve A’s velocity into components.
The angle is measured north of east.
The horizontal component is
\[ v_{A/G,x}=10.0\cos37^\circ =8.0\\,\mathrm{m\\,s^{-1}} \]
The vertical component is
\[ v_{A/G,y}=10.0\sin37^\circ =6.0\\,\mathrm{m\\,s^{-1}} \]
Therefore,
\[ \vec v_{A/G}=(8.0,6.0) \]
B travels directly east, so
\[ \vec v_{B/G}=(4.0,0) \]
All components are in \(\mathrm{m\\,s^{-1}}\).
Step 3: Subtract the observer’s velocity.
\[ \begin{aligned} \vec v_{A/B} &=(8.0,6.0)-(4.0,0)\\\ &=(4.0,6.0)\\,\mathrm{m\\,s^{-1}} \end{aligned} \]
Both components are positive, so A moves northeast relative to B.
Step 4: Calculate the relative speed.
Using Pythagoras:
\[ \begin{aligned} s_{A/B} &=\sqrt{4.0^2+6.0^2}\\\ &=\sqrt{52}\\\ &=7.21\\,\mathrm{m\\,s^{-1}} \end{aligned} \]
Step 5: Calculate the direction.
Measure the angle north of east:
\[ \begin{aligned} \theta &=\tan^{-1}\left(\frac{6.0}{4.0}\right)\\\ &=56.3^\circ \end{aligned} \]
The final answer is
\[ \boxed{ 7.21\\,\mathrm{m\\,s^{-1}}, \quad 56.3^\circ \text{ north of east} } \]
A passenger on B measures A moving northeast at 7.21 m/s, even though A’s ground-frame velocity points at a different angle.
Worked example: A boat crossing a moving river
A boat travels north at 4.0 m/s relative to the water. The river current flows east at 3.0 m/s relative to the riverbank.
The river is 120 m wide. Calculate:
- The boat’s velocity relative to the riverbank.
- The time required to reach the opposite bank.
- The boat’s downstream displacement.
Assume the boat maintains constant velocity relative to the water, and the current is uniform.
Step 1: Identify the reference frames.
This time, we have three frames:
- B: boat.
- W: water.
- G: riverbank.
The required velocity is the boat’s velocity relative to the bank.
The correct vector relationship is
\[ \vec v_{B/G} = \vec v_{B/W}+\vec v_{W/G} \]
Notice that we are adding two velocities because the boat moves relative to the water, and the water moves relative to the bank.
Step 2: Resolve the velocities into components.
Choose east as positive \(x\) and north as positive \(y\).
The boat’s velocity relative to the water is
\[ \vec v_{B/W}=(0,4.0) \]
The current’s velocity relative to the bank is
\[ \vec v_{W/G}=(3.0,0) \]
Therefore,
\[ \begin{aligned} \vec v_{B/G} &=(0,4.0)+(3.0,0)\\\ &=(3.0,4.0)\\,\mathrm{m\\,s^{-1}} \end{aligned} \]
Step 3: Calculate the boat’s ground-frame speed.
\[ \begin{aligned} s_{B/G} &=\sqrt{3.0^2+4.0^2}\\\ &=5.0\\,\mathrm{m\\,s^{-1}} \end{aligned} \]
Its direction is
\[ \begin{aligned} \theta &=\tan^{-1}\left(\frac{3.0}{4.0}\right)\\\ &=36.9^\circ \end{aligned} \]
The boat travels at 5.0 m/s, 36.9° east of north relative to the bank.
Step 4: Calculate the crossing time.
Here’s an important detail.
The river is 120 m wide in the northward direction. The boat’s northward velocity component is 4.0 m/s, not its diagonal ground speed of 5.0 m/s.
Therefore,
\[ \begin{aligned} t &=\frac{\text{northward distance}} {\text{northward velocity}}\\\ &=\frac{120}{4.0}\\\ &=30\\,\mathrm s \end{aligned} \]
Step 5: Calculate the downstream displacement.
The current carries the boat east at 3.0 m/s for 30 seconds.
\[ \begin{aligned} d_{\text{east}} &=v_{\text{current}}t\\\ &=3.0\times30\\\ &=90\\,\mathrm m \end{aligned} \]
The boat reaches the opposite bank 90 m east of the point directly across from its starting position.
This example introduces an important extension of the triangle method: you can add velocities across several reference frames, provided the frames are connected in the correct order.
## Questions and solutions
### Question 1
Two skaters move across a level skating rink. An observer standing beside the rink measures their velocities:
- Skater A travels east at \(9.0\,\mathrm{m\,s^{-1}}\).
- Skater B travels north at \(12.0\,\mathrm{m\,s^{-1}}\).
Calculate the velocity of A relative to B, including its magnitude and direction.
[IMAGE: Overhead view of two skaters with their ground-frame velocity arrows shown from a common origin. A points east at 9.0 m/s and B points north at 12.0 m/s. Do not include the relative-velocity arrow or its direction.]
### Solution 1
The velocity of A relative to B is \(15.0\,\mathrm{m\,s^{-1}}\), directed \(53.1^\circ\) south of east.
Start with the relative-velocity equation:
\[
\vec v_{A/B}
=
\vec v_{A/G}-\vec v_{B/G}
\]
**Step 1**: Resolve the given velocities into components.
Choose east as positive \(x\) and north as positive \(y\).
\[
\vec v_{A/G}=(9.0,0)
\]
\[
\vec v_{B/G}=(0,12.0)
\]
**Step 2**: Subtract the observer's velocity.
\[
\begin{aligned}
\vec v_{A/B}
&=(9.0,0)-(0,12.0)\\
&=(9.0,-12.0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The negative \(y\)-component indicates south.
**Step 3**: Calculate the magnitude.
\[
\begin{aligned}
s_{A/B}
&=\sqrt{9.0^2+12.0^2}\\
&=\sqrt{225}\\
&=15.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
**Step 4**: Calculate the direction.
\[
\begin{aligned}
\theta
&=\tan^{-1}\left(\frac{12.0}{9.0}\right)\\
&=53.1^\circ
\end{aligned}
\]
[IMAGE: Completed relative-velocity triangle for Question 1. Show A's 9 m/s east vector and B's 12 m/s north vector from a common origin. Join B's arrowhead to A's arrowhead with a 15 m/s southeast arrow. Label the 53.1 degree angle south of east.]
The relative velocity is therefore
\[
\boxed{
15.0\,\mathrm{m\,s^{-1}},
\quad 53.1^\circ
\text{ south of east}
}
\]
The important point is that we subtract the observer's velocity as a vector. Simply subtracting the two speeds would give \(3.0\,\mathrm{m\,s^{-1}}\), which is incorrect because the skaters travel in perpendicular directions.
### Question 2
A boat must cross a river and arrive at a point directly opposite its starting position.
The river current flows east at \(2.4\,\mathrm{m\,s^{-1}}\) relative to the riverbank. The boat travels at \(4.0\,\mathrm{m\,s^{-1}}\) relative to the water.
The river is 96 m wide.
Determine:
a) The direction in which the boat must head relative to the water.
b) The boat's velocity relative to the riverbank.
c) The time required to cross the river.
Assume the current is uniform and the boat maintains constant speed relative to the water.
[IMAGE: Top-down river-crossing question setup. Show parallel north and south banks, a 96 m river width, the boat starting on the southern bank, and an eastward current arrow labelled 2.4 m/s. Mark the destination directly north of the starting point. State that the boat's speed relative to the water is 4.0 m/s, but do not draw its heading or the completed velocity triangle.]
### Solution 2
The boat must head \(36.9^\circ\) west of north. Its ground-frame velocity is \(3.2\,\mathrm{m\,s^{-1}}\) north, giving a crossing time of \(30.0\) seconds.
**Step 1**: Identify the reference frames.
Let:
- B represent the boat.
- W represent the water.
- G represent the riverbank.
The velocity relationship is
\[
\vec v_{B/G}
=
\vec v_{B/W}+\vec v_{W/G}
\]
**Step 2**: Determine the required horizontal component.
The boat must arrive directly opposite its starting position.
Therefore, its east-west ground-frame velocity must be zero.
The current contributes \(+2.4\,\mathrm{m\,s^{-1}}\) east.
The boat must therefore have a westward component of
\[
v_{B/W,x}=-2.4\,\mathrm{m\,s^{-1}}
\]
This is the key step. The boat must aim upstream to cancel the current.
**Step 3**: Calculate the northward component of the boat's velocity relative to the water.
The boat's speed relative to the water is \(4.0\,\mathrm{m\,s^{-1}}\).
Using Pythagoras:
\[
\begin{aligned}
v_{B/W,y}
&=\sqrt{4.0^2-2.4^2}\\
&=\sqrt{16-5.76}\\
&=3.2\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
**Step 4**: Calculate the heading angle.
Measure the angle west of north:
\[
\begin{aligned}
\theta
&=\sin^{-1}\left(\frac{2.4}{4.0}\right)\\
&=36.9^\circ
\end{aligned}
\]
The boat must head \(36.9^\circ\) west of north.
[IMAGE: Completed boat velocity triangle. Show the boat's velocity relative to water as a 4.0 m/s arrow pointing northwest, with components 2.4 m/s west and 3.2 m/s north. Add the 2.4 m/s eastward current vector head-to-tail. Show the resultant ground-frame velocity pointing directly north at 3.2 m/s. Label the 36.9 degree heading angle.]
**Step 5**: Calculate the velocity relative to the riverbank.
The horizontal components cancel:
\[
\begin{aligned}
v_{B/G,x}&=-2.4+2.4=0\\
v_{B/G,y}&=3.2+0=3.2
\end{aligned}
\]
Therefore,
\[
\vec v_{B/G}
=
(0,3.2)\,\mathrm{m\,s^{-1}}
\]
**Step 6**: Calculate the crossing time.
Use the northward velocity component:
\[
\begin{aligned}
t
&=\frac{96}{3.2}\\
&=30.0\,\mathrm s
\end{aligned}
\]
The boat reaches the opposite bank after 30.0 seconds.
Notice that the boat travels at 4.0 m/s relative to the water but only 3.2 m/s relative to the bank. The two measurements describe different reference frames.
### Question 3
Rain falls vertically downwards at \(9.0\,\mathrm{m\,s^{-1}}\) relative to the ground.
Two cyclists travel along the same horizontal road:
- Cyclist A travels east at \(12.0\,\mathrm{m\,s^{-1}}\).
- Cyclist B travels west at \(12.0\,\mathrm{m\,s^{-1}}\).
Calculate the velocity of the rain relative to each cyclist, including the relative speed and direction.
Explain why the cyclists observe the rain moving in opposite directions, despite travelling at the same speed.
[IMAGE: Side-by-side overhead and side-view composite showing two cyclists travelling in opposite horizontal directions, with vertical downward rain arrows in the stationary ground frame. Label cyclist A's eastward velocity as 12 m/s, cyclist B's westward velocity as 12 m/s, and the rain's downward velocity as 9 m/s. Do not show either apparent rain direction.]
### Solution 3
Both cyclists measure a relative rain speed of \(15.0\,\mathrm{m\,s^{-1}}\), but the apparent rain directions are opposite.
The reason is that the cyclists have opposite ground-frame velocity vectors.
**Step 1**: Establish the coordinate system.
Choose east as positive \(x\) and upward as positive \(y\).
The rain falls vertically downwards:
\[
\vec v_{R/G}=(0,-9.0)
\]
Cyclist A travels east:
\[
\vec v_{A/G}=(12.0,0)
\]
Cyclist B travels west:
\[
\vec v_{B/G}=(-12.0,0)
\]
All velocities are in \(\mathrm{m\,s^{-1}}\).
**Step 2**: Calculate the rain's velocity relative to cyclist A.
\[
\begin{aligned}
\vec v_{R/A}
&=\vec v_{R/G}-\vec v_{A/G}\\
&=(0,-9.0)-(12.0,0)\\
&=(-12.0,-9.0)
\end{aligned}
\]
The rain appears to move west and downwards.
Its relative speed is
\[
\begin{aligned}
s_{R/A}
&=\sqrt{(-12.0)^2+(-9.0)^2}\\
&=\sqrt{225}\\
&=15.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The angle west of the downward vertical is
\[
\begin{aligned}
\theta
&=\tan^{-1}\left(\frac{12.0}{9.0}\right)\\
&=53.1^\circ
\end{aligned}
\]
**Step 3**: Calculate the rain's velocity relative to cyclist B.
\[
\begin{aligned}
\vec v_{R/B}
&=\vec v_{R/G}-\vec v_{B/G}\\
&=(0,-9.0)-(-12.0,0)\\
&=(12.0,-9.0)
\end{aligned}
\]
The rain appears to move east and downwards.
Its relative speed is
\[
\begin{aligned}
s_{R/B}
&=\sqrt{12.0^2+(-9.0)^2}\\
&=15.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The angle east of the downward vertical is also
\[
\theta=53.1^\circ
\]
[IMAGE: Two completed relative-velocity triangles side by side. For cyclist A, show the rain's downward ground velocity and the cyclist's eastward velocity, producing a southwest relative rain vector. For cyclist B, show the same downward rain velocity and a westward cyclist velocity, producing a southeast relative rain vector. Label both relative speeds as 15 m/s and both angles as 53.1 degrees from downward vertical.]
The results are:
| Observer | Relative rain velocity | Relative speed |
|---|---|---:|
| Cyclist A | Southwest, 53.1° west of down | 15.0 m/s |
| Cyclist B | Southeast, 53.1° east of down | 15.0 m/s |
The ground-frame rain velocity is identical for both cyclists. However, each cyclist subtracts a different observer velocity.
This produces equal relative speeds but opposite apparent directions.
It is another example of why a speed reading alone cannot tell you the direction of relative motion.
### Question 4
A passenger travels east on a train at \(4.0\,\mathrm{m\,s^{-1}}\) relative to the ground.
The passenger observes a drone moving at \(10.0\,\mathrm{m\,s^{-1}}\), at \(37^\circ\) north of west, relative to the train.
Calculate the drone's velocity relative to the ground, including its speed and direction.
[IMAGE: Side-view and overhead composite of an eastbound train with a passenger observing a drone. Show the train's ground-frame velocity as 4.0 m/s east. Show the drone's measured velocity relative to the train as 10.0 m/s at 37 degrees north of west. Do not draw the drone's ground-frame velocity.]
### Solution 4
The drone travels at approximately \(7.2\,\mathrm{m\,s^{-1}}\), \(56.3^\circ\) north of west relative to the ground.
This question reverses the usual calculation. We are given the object's relative velocity and the observer's ground velocity, and must determine the object's ground velocity.
**Step 1**: Identify the reference frames.
Let A represent the drone, B represent the train, and G represent the ground.
We know:
\[
\vec v_{A/B}
\]
and
\[
\vec v_{B/G}
\]
We need:
\[
\vec v_{A/G}
\]
Rearrange the relative-velocity equation:
\[
\boxed{
\vec v_{A/G}
=
\vec v_{A/B}+\vec v_{B/G}
}
\]
**Step 2**: Resolve the drone's relative velocity into components.
The drone moves \(37^\circ\) north of west.
Therefore, its horizontal component is negative and its vertical component is positive.
\[
v_{A/B,x}=-10.0\cos37^\circ=-8.0
\]
\[
v_{A/B,y}=10.0\sin37^\circ=6.0
\]
Thus,
\[
\vec v_{A/B}=(-8.0,6.0)
\]
**Step 3**: Write the train's ground-frame velocity.
The train moves east:
\[
\vec v_{B/G}=(4.0,0)
\]
**Step 4**: Add the vectors.
\[
\begin{aligned}
\vec v_{A/G}
&=(-8.0,6.0)+(4.0,0)\\
&=(-4.0,6.0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The drone's ground-frame velocity points northwest.
**Step 5**: Calculate its magnitude.
\[
\begin{aligned}
s_{A/G}
&=\sqrt{(-4.0)^2+6.0^2}\\
&=\sqrt{52}\\
&=7.21\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
**Step 6**: Calculate its direction.
\[
\begin{aligned}
\theta
&=\tan^{-1}\left(\frac{6.0}{4.0}\right)\\
&=56.3^\circ
\end{aligned}
\]
[IMAGE: Completed head-to-tail triangle for Question 4. Begin with the train's 4.0 m/s eastward ground velocity. From its arrowhead, draw the drone's 10.0 m/s relative velocity at 37 degrees north of west. Draw the resultant from the original starting point to the final endpoint, pointing northwest with components 4.0 m/s west and 6.0 m/s north.]
The final result is
\[
\boxed{
7.21\,\mathrm{m\,s^{-1}},
\quad 56.3^\circ
\text{ north of west}
}
\]
The important distinction is that we add the train's ground velocity because the drone's velocity relative to the train is already known.
Using subtraction here would reverse the reference-frame relationship.
### Question 5
Two rescue boats, A and B, travel through a wide river.
At \(t=0\), boat A is 60 m west and 20 m north of boat B.
Their velocities relative to the water remain constant:
- Boat A travels east at \(5.0\,\mathrm{m\,s^{-1}}\).
- Boat B travels north at \(3.0\,\mathrm{m\,s^{-1}}\).
The river current is spatially uniform and affects both boats equally. It initially flows east at \(2.0\,\mathrm{m\,s^{-1}}\) relative to the riverbank.
After 4.0 seconds, the current changes direction instantly and flows west at \(1.0\,\mathrm{m\,s^{-1}}\).
Assume both boats maintain their stated velocities relative to the water throughout the motion.
Answer the following:
a) Calculate the ground-frame velocity of each boat before and after the current changes.
b) Determine whether the boats collide at any time after \(t=0\).
c) If they do not collide, calculate when their east-west coordinates coincide and their north-south separation at that instant.
d) Explain why the change in current does or does not change their relative velocity.
[IMAGE: Overhead river diagram showing boats A and B at their initial positions, with A 60 m west and 20 m north of B. Show A's eastward water-relative velocity of 5.0 m/s, B's northward water-relative velocity of 3.0 m/s, and the eastward current before 4.0 seconds. Include a separate indication that the current subsequently reverses west. Do not show the relative-velocity vector or collision conclusion.]
### Solution 5
The boats do not collide. Their relative velocity remains \(5.0\,\mathrm{m\,s^{-1}}\) east and \(3.0\,\mathrm{m\,s^{-1}}\) south, even when the current reverses.
This is the most important idea in the question: a shared current can change both boats' ground-frame velocities without changing their relative velocity.
**Step 1**: Calculate the ground-frame velocities before the current changes.
Choose east as positive \(x\) and north as positive \(y\).
For boat A:
\[
\vec v_{A/W}=(5.0,0)
\]
For boat B:
\[
\vec v_{B/W}=(0,3.0)
\]
The current is
\[
\vec v_{W/G}=(2.0,0)
\]
Using velocity addition:
\[
\begin{aligned}
\vec v_{A/G}
&=(5.0,0)+(2.0,0)\\
&=(7.0,0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
For boat B:
\[
\begin{aligned}
\vec v_{B/G}
&=(0,3.0)+(2.0,0)\\
&=(2.0,3.0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
**Step 2**: Calculate the ground-frame velocities after the current reverses.
The new current velocity is
\[
\vec v_{W/G}=(-1.0,0)
\]
For boat A:
\[
\begin{aligned}
\vec v_{A/G}
&=(5.0,0)+(-1.0,0)\\
&=(4.0,0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
For boat B:
\[
\begin{aligned}
\vec v_{B/G}
&=(0,3.0)+(-1.0,0)\\
&=(-1.0,3.0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
Both ground-frame velocities have changed.
Now comes the important part.
**Step 3**: Calculate the relative velocity before the current changes.
\[
\begin{aligned}
\vec v_{A/B}
&=\vec v_{A/G}-\vec v_{B/G}\\
&=(7.0,0)-(2.0,3.0)\\
&=(5.0,-3.0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
A moves east and south relative to B.
**Step 4**: Calculate the relative velocity after the current changes.
\[
\begin{aligned}
\vec v_{A/B}
&=(4.0,0)-(-1.0,3.0)\\
&=(5.0,-3.0)\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The relative velocity has not changed.
[IMAGE: Side-by-side comparison of the two boats' ground-frame velocity triangles before and after the current reversal. Before: A moves east at 7 m/s and B moves northeast with components 2 m/s east and 3 m/s north. After: A moves east at 4 m/s and B moves northwest with components 1 m/s west and 3 m/s north. In both cases, show the identical relative vector from B's arrowhead to A's arrowhead: 5 m/s east and 3 m/s south. Make the cancellation of the shared current visually clear.]
**Step 5**: Determine whether the boats collide.
The initial position of A relative to B is
\[
\vec r_{A/B}(0)=(-60,20)\,\mathrm m
\]
The relative velocity remains constant:
\[
\vec v_{A/B}=(5,-3)\,\mathrm{m\,s^{-1}}
\]
Therefore, the relative position after time \(t\) is
\[
\vec r_{A/B}(t)
=
(-60+5t,\ 20-3t)
\]
For a collision, both components must equal zero at the same instant.
First, consider the east-west component:
\[
-60+5t=0
\]
\[
t=12\,\mathrm s
\]
Now consider the north-south component:
\[
20-3t=0
\]
\[
t=\frac{20}{3}
=6.67\,\mathrm s
\]
These times are different.
Therefore, the boats cannot occupy the same position at the same time.
They do not collide.
**Step 6**: Calculate their north-south separation when their east-west coordinates coincide.
Their east-west coordinates coincide after 12 seconds.
Substitute \(t=12\) into the north-south component:
\[
\begin{aligned}
y_{A/B}
&=20-3(12)\\
&=20-36\\
&=-16\,\mathrm m
\end{aligned}
\]
The negative sign indicates that A is south of B.
At the instant their east-west coordinates match, boat A is 16 m south of boat B.
**Step 7**: Explain why the current reversal does not change their relative velocity.
The key is that both boats experience the same current.
Before the reversal:
\[
\vec v_{A/G}
=
\vec v_{A/W}+\vec v_{W/G}
\]
\[
\vec v_{B/G}
=
\vec v_{B/W}+\vec v_{W/G}
\]
Subtracting:
\[
\begin{aligned}
\vec v_{A/B}
&=\vec v_{A/G}-\vec v_{B/G}\\
&=(\vec v_{A/W}+\vec v_{W/G})\\
&\quad-(\vec v_{B/W}+\vec v_{W/G})\\
&=\vec v_{A/W}-\vec v_{B/W}
\end{aligned}
\]
The common current cancels.
The same cancellation happens after the reversal, because the new current is still shared by both boats.
This result holds even if the common current changes with time, provided both boats experience the same current at each instant. If the current differs at their respective locations, the cancellation is no longer guaranteed.
The tempting mistake is to assume that a change in the ground-frame velocities must change the relative velocity. It doesn't necessarily do so. What matters is the difference between the two velocities, not their individual values.
## 7. The final arrow check
Before submitting a relative-velocity question, check these five things:
- Have you clearly identified the object and observer?
- Are the original velocities measured in the same reference frame?
- Have you used vector subtraction rather than subtracting scalar speeds?
- Does the relative-velocity arrow point from the observer's arrowhead to the object's arrowhead?
- Does the magnitude and direction make physical sense?
If you remember only one drawing rule, remember this:
**For \(\vec v_{A/B}\), draw the arrow from the tip of \(\vec v_{B/G}\) to the tip of \(\vec v_{A/G}\).**
That direction comes directly from the equation, so you don't need to guess or reverse arrows twice.
For further practice with the difference between speed and velocity, see [Can Two Objects Have the Same Speed but Different Velocities?](https://tutorgum.com/learn/resources/hsc-physics-can-two-objects-have-the-same-speed-but-a-large-relative-velocity/).
Once your relative-velocity triangle is correct, the next step is using its components to analyse relative displacement, interception, and whether two moving objects will actually meet.