Animated Solution for Physics - Kinematics: Consider two steamers A and B on a calm sea. Steamer A is moving towards the north with a constant speed vA=30 km/h and steamer B towards the south with a constant speed vB=10 km/h. If smoke ejected by steamer A spreads in a straight line from the steamer towards the west and smoke ejected by steamer B spreads in another straight line from the steamer towards the north-west, determine magnitude and direction of the wind velocity.
Visualized Solution
\text{Coordinate System & Velocities}
Let East be +x (i^) and North be +y (j^).
vA=30j^
vB=−10j^
Concept of Smoke Direction
The smoke spreads in the direction of the wind relative to the steamer.
vwA=vw−vA
vwB=vw−vB
Analyzing Steamer A
Let wind velocity be vw=vxi^+vyj^.
vwA=(vxi^+vyj^)−30j^
vwA=vxi^+(vy−30)j^
Wind’s North-South Component
Smoke from A goes West (−i^).
Therefore, the y-component of vwA must be zero.
vy−30=0⟹vy=30 km/h
Analyzing Steamer B
vwB=(vxi^+30j^)−(−10j^)
vwB=vxi^+40j^
Wind’s East-West Component
Smoke from B goes North-West (−i^+j^).
For North-West, the x and y components must be equal in magnitude but opposite in sign.
∣vx∣=40⟹vx=−40 km/h
Final Wind Velocity
vw=−40i^+30j^
Magnitude: ∣vw∣=(−40)2+302=50 km/h
Direction: tanθ=3040=34
θ≈53∘ West of North
The Way Forward
What if the sea wasn't calm and had a water current?
How would the relative velocity of the steamers change?
00:00 / 00:00
The Sigma Insight: Relative Velocity
Solution Diagram
The problem of the two steamers and their smoke trails is a classic, elegant puzzle in kinematics. It beautifully bridges the gap between abstract vector algebra and a highly visual, physical phenomenon. When you stand on a moving boat and watch the smoke billow from its chimney, you are not seeing the true velocity of the wind. Instead, you are witnessing a profound illusion created by your own motion. The smoke reveals the wind's velocity relative to you.
Let's embark on a journey to decode this illusion and uncover the true speed and direction of the wind sweeping across our calm sea.
The Illusion of Smoke
Imagine you are the captain of Steamer A, cruising steadily towards the North at 30 km/h. You look up at the smokestack and see the smoke trailing perfectly towards the West. Your intuition might scream, "Ah, the wind is blowing West!" But wait. If you were standing still on the calm sea, the smoke would indeed follow the true wind. However, because you are slicing through the air towards the North, you are creating an "apparent" wind blowing towards the South. The smoke you see is being pushed by a combination of the true wind and this apparent wind.
In physics, we formalize this using the concept of relative velocity. The velocity of the smoke as seen from the steamer is exactly the velocity of the wind relative to the steamer. Mathematically, this is expressed as:
vwA=vw−vA
Where vwA is the relative velocity of the wind with respect to Steamer A, vw is the absolute velocity of the wind, and vA is the absolute velocity of Steamer A.
Setting Up the Mathematical Stage
To solve this elegantly, we must anchor our physical situation to a rigid mathematical coordinate system. Let's define the East direction as the positive x-axis (+i^) and the North direction as the positive y-axis (+j^).
Given this framework, we can write down the velocity vectors for our two steamers:
- Steamer A is moving North at 30 km/h: vA=30j^
- Steamer B is moving South at 10 km/h: vB=−10j^
The true wind velocity is the mystery we need to solve. Let's represent it with unknown components:
vw=vxi^+vyj^
Our goal is to find the exact values of vx and vy.
Decoding Steamer A's Smoke
Let's apply our relative velocity master equation to Steamer A.
vwA=(vxi^+vyj^)−30j^
Grouping the components together, we get:
vwA=vxi^+(vy−30)j^
Now, we look at the crucial clue provided in the problem: the smoke from Steamer A spreads in a straight line towards the West. In our coordinate system, West corresponds to the negative x-direction (−i^).
If a vector points purely West, it cannot have any North or South component. This means the y-component of our relative velocity vector vwA must be exactly zero!
vy−30=0
vy=30 km/h
Just like that, by analyzing the lack of North-South drift in Steamer A's smoke, we have uncovered the North-South component of the true wind. The wind is blowing Northwards at 30 km/h (in addition to whatever its East-West motion is).
Decoding Steamer B's Smoke
Now, let's shift our perspective to Steamer B. We already know the y-component of the wind, so we can update our wind vector:
vw=vxi^+30j^
Applying the relative velocity equation for Steamer B:
vwB=vw−vB
vwB=(vxi^+30j^)−(−10j^)
Notice the double negative! Because Steamer B is moving South, subtracting its velocity effectively adds a Northward component to the relative wind.
vwB=vxi^+40j^
The problem states that the smoke from Steamer B spreads towards the North-West. The North-West direction is a perfect diagonal, exactly bisecting North and West. For a vector to point North-West, its Westward component and its Northward component must be perfectly balanced in magnitude.
Since the Northward (y) component is 40 km/h, the Westward (x) component must also have a magnitude of 40 km/h. Because it's pointing West, the value must be negative:
vx=−40 km/h
The Final Vector Assembly
We have successfully decoded both clues and found the components of the true wind velocity:
vw=−40i^+30j^
To find the magnitude (speed) of the wind, we apply the Pythagorean theorem to its components:
∣vw∣=(−40)2+(30)2
∣vw∣=1600+900=2500=50 km/h
Finally, we need to determine the direction. We can find the angle θ the wind makes with the North direction (the y-axis).
tanθ=∣y-component∣∣x-component∣=3040=34
θ=tan−1(34)≈53∘
Since the x-component is negative (West) and the y-component is positive (North), the wind is blowing at an angle of approximately 53∘ West of North.
This problem is a fantastic demonstration of how independent observations from different moving reference frames can be synthesized to reconstruct the absolute truth of a physical system. The next time you see smoke trailing from a moving vehicle, remember: you are looking at a vector subtraction problem playing out in the real world!