5.3 Upstream, Downstream Boat Velocity, Still Water Speed & Circular Track Races

Kinematic Mechanics of Boats and Streams

When a vessel moves in a fluid medium, its net ground speed is the vector sum of its propulsion velocity and the fluid's velocity.

DOWNSTREAM MOTION (Vessel moving WITH the stream current):
   ──► Boat Speed u        ──► Stream Speed v
   Net Ground Speed D = u + v

UPSTREAM MOTION (Vessel moving AGAINST the stream current):
   ◄── Boat Speed u        ──► Stream Speed v
   Net Ground Speed U = u - v   (Constraint: u > v for forward progress)

Primary Vector Decompositions

Let:

• $u =$ Speed of boat in still water
• $v =$ Speed of stream (river current)
• $D =$ Downstream speed (motion in direction of river flow)
• $U =$ Upstream speed (motion against direction of river flow)

From the physical definitions:

$$D = u + v \quad \text{--- (1)}$$ $$U = u - v \quad \text{--- (2)}$$

Adding (1) and (2) yields the still water speed $u$:

$$D + U = 2u \implies u = \frac{D + U}{2}$$

Subtracting (2) from (1) yields the stream current speed $v$:

$$D - U = 2v \implies v = \frac{D - U}{2}$$

Ratio of Upstream and Downstream Times for Equal Distance

For a fixed distance $d$, let $T_u$ be upstream time and $T_d$ be downstream time:

$$T_u = \frac{d}{u - v}, \quad T_d = \frac{d}{u + v}$$ $$\frac{T_u}{T_d} = \frac{u + v}{u - v}$$

The Time Multiplier Shortcut ($n$ times as long upstream)

If an exam question states that rowing upstream takes $n$ times as long as rowing downstream for the same distance ($T_u = n \cdot T_d$):

$$\frac{u + v}{u - v} = n \implies u + v = n(u - v) \implies u(n - 1) = v(n + 1)$$ $$u = v \left(\frac{n + 1}{n - 1}\right)$$

The Round-Trip Stream Velocity Trap

Suppose a boat rows distance $d$ downstream and returns the same distance $d$ upstream:

• Total Distance $= 2d$
• Total Time $= T_d + T_u = \frac{d}{u + v} + \frac{d}{u - v} = d \left(\frac{(u - v) + (u + v)}{u^2 - v^2}\right) = \frac{2du}{u^2 - v^2}$
• Average Speed:

$$V_{\text{avg}} = \frac{2d}{\frac{2du}{u^2 - v^2}} = \frac{u^2 - v^2}{u} = u \left(1 - \frac{v^2}{u^2}\right)$$

Relative Speed of Two Craft in a River

When boats with still-water speeds $u_1,u_2$ move toward each other, one downstream and one upstream, through the same uniform current $v$ (with $u_2>v$):

$$\text{Relative Approach Speed} = (u_1 + v) + (u_2 - v) = u_1 + u_2$$

The stream velocity cancels out completely! The closing speed depends exclusively on their still water capabilities.

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Linear Races, Handicaps & Dead Heats

In competitive sprint and distance races along a straight track of length $L$:

Start Line                                                      Finish Line
   │                                                                 │
   ├── Runner A (Starts at 0 m) ────────────────────────────────────►│
   │                                                                 │
   │◄── Start x m ──►                                                │
   │                 ├── Runner B (Starts at x m) ──────────────────►│
   │                                                                 │

1. Distance Handicap

"A gives B a start of $x$ meters":

• B is positioned $x$ meters ahead of A at the starting signal.
• A must run the full race course $L$.
• B needs to cover only $(L - x)$ meters.
• If A beats B by $y$ meters, then while A covers $L$ meters, B covers only $(L - x - y)$ meters.
• Speed ratio: $\frac{S_A}{S_B} = \frac{L}{L - x - y}$.

2. Time Handicap

"A gives B a start of $t$ seconds":

• B begins running $t$ seconds before A starts.
• If A completes the race in $T_A$ seconds, B has been running for $(T_A + t)$ seconds.

3. The Golden Rule of Races

$$S_B = \frac{x}{t}$$

4. Dead Heat Condition

A race ends in a dead heat when finish instants are identical. With simultaneous starts, their elapsed race times are equal; with a time handicap, running durations can differ even though the finish instants match.

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Circular Track Races: Periodic Concurrency & Rendezvous

Closed circular tracks of circumference $L$ introduce periodic topology, where faster runners lap slower runners.

                       12 o'clock (Start Line)
                              ▲
                         . ───┴─── .
                      '       │       '
                    /         │         \
                   │          │          │   Track Circumference = L
                   │          O          │   Runner A: Speed u
                    \                   /    Runner B: Speed v (u > v)
                      .               .
                         ' . ___ . '

1. Time to Meet for the First Time Anywhere on the Track

• Opposite Directions: The two runners run towards each other. Their relative approach speed is $u + v$. The sum of their distances covered must equal one circumference $L$:

$$T_{\text{meet, opp}} = \frac{L}{u + v}$$

• Same Direction: The faster runner must gain a lead of exactly one full lap ($L$) over the slower runner. Their relative separation speed is $u - v$:

$$T_{\text{meet, same}} = \frac{L}{u - v}$$

2. Time to Meet for the First Time at the STARTING Point

Both runners must be at the original starting point simultaneously:

• Time for A to complete 1 full lap: $t_A = \frac{L}{u}$
• Time for B to complete 1 full lap: $t_B = \frac{L}{v}$
• A arrives at the start line at multiples of $t_A$: $\{t_A, 2t_A, 3t_A, \dots\}$
• B arrives at the start line at multiples of $t_B$: $\{t_B, 2t_B, 3t_B, \dots\}$

The first time they coincide at the start line is the Lowest Common Multiple (LCM):

$$T_{\text{start}} = \text{LCM}\left(t_A, t_B\right) = \text{LCM}\left(\frac{L}{u}, \frac{L}{v}\right)$$ $$\text{LCM}\left(\frac{p_1}{q_1}, \frac{p_2}{q_2}\right) = \frac{\text{LCM}(p_1, p_2)}{\text{HCF}(q_1, q_2)}$$

3. Number of Distinct Meeting Points on the Track

Reduce the ratio of their speeds to lowest irreducible integer terms:

$$\frac{u}{v} = \frac{a}{b}, \quad \text{where } \gcd(a, b) = 1$$

• Running in the SAME direction:

$$\text{Number of distinct meeting points} = a - b$$

These points divide the circumference into $(a - b)$ equal arcs.

• Running in OPPOSITE directions:

$$\text{Number of distinct meeting points} = a + b$$

These points divide the circumference into $(a + b)$ equal arcs.

4. Extension to Three Runners on a Circular Track

For three runners A, B, and C with lap times $t_A, t_B, t_C$:

• First meeting at the starting point: $\text{LCM}(t_A, t_B, t_C)$
• First meeting anywhere on track: Find first meeting time between A and B ($t_{AB}$), and between B and C ($t_{BC}$). Then $T = \text{LCM}(t_{AB}, t_{BC})$.

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Assumptions behind the circular formulas

These formulas assume simultaneous departure from the same point, constant speeds and no stops. “First meeting” means the first meeting after departure. The same-direction lap formula assumes unequal speeds; equal-speed runners starting together remain together. LCM-based return formulas require commensurable lap times, as in the rational-number examples here. With different starting positions, first account for the initial separation rather than applying a full-lap formula directly.

Worked Examples

Worked Example 1: Simultaneous Upstream/Downstream Linear System

Problem: A motorized railway inspection boat operating along a river corridor travels $44 \text{ km}$ downstream and $30 \text{ km}$ upstream in a total time of $10 \text{ hours}$. In another inspection trial, it travels $55 \text{ km}$ downstream and $40 \text{ km}$ upstream in $13 \text{ hours}$. Determine:

1. The operational speed of the inspection boat in still water.
2. The velocity of the river current.

• Step 1: Set up the simultaneous equations: Let downstream speed be $D = u + v$ and upstream speed be $U = u - v$.

$$\frac{44}{D} + \frac{30}{U} = 10 \quad \text{--- (Equation 1)}$$ $$\frac{55}{D} + \frac{40}{U} = 13 \quad \text{--- (Equation 2)}$$

• Step 2: Substitute reciprocal variables: Let $x = \frac{1}{D}$ and $y = \frac{1}{U}$:

$$44x + 30y = 10 \quad \text{--- (1)}$$ $$55x + 40y = 13 \quad \text{--- (2)}$$

• Step 3: Eliminate $y$ by matching coefficients: Multiply (1) by 4: $176x + 120y = 40$ Multiply (2) by 3: $165x + 120y = 39$ Subtract the two equations:

$$(176x - 165x) = 40 - 39 \implies 11x = 1 \implies x = \frac{1}{11}$$

Since $x = \frac{1}{D} = \frac{1}{11}$, Downstream Speed $D = 11 \text{ km/h}$.

• Step 4: Solve for $y$ and $U$: Substitute $x = \frac{1}{11}$ into (1):

$$44 \left(\frac{1}{11}\right) + 30y = 10 \implies 4 + 30y = 10 \implies 30y = 6 \implies y = \frac{6}{30} = \frac{1}{5}$$

Since $y = \frac{1}{U} = \frac{1}{5}$, Upstream Speed $U = 5 \text{ km/h}$.

• Step 5: Decompose into still water speed $u$ and current $v$:

$$u = \frac{D + U}{2} = \frac{11 + 5}{2} = \frac{16}{2} = 8 \text{ km/h}$$ $$v = \frac{D - U}{2} = \frac{11 - 5}{2} = \frac{6}{2} = 3 \text{ km/h}$$

• Conclusion: The boat speed in still water is 8 km/h, and the river current is 3 km/h.

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Worked Example 2: Linear Race with Dual Handicaps

Problem: In a $1000\text{-meter}$ sprint trial at the Railway Sports Promotion Board athletics academy, runner A can give runner B a start of $50 \text{ meters}$ and still beat him by $10 \text{ seconds}$. Alternatively, if runner A gives runner B a start of $20 \text{ seconds}$, runner B beats runner A by $50 \text{ meters}$. Determine the running speeds of runner A and runner B in meters per second.

• Step 1: Formulate the kinematic equations: Let the speeds of runners A and B be $S_A$ and $S_B$.
  • Scenario 1: A runs $1000 \text{ m}$, B runs $1000 - 50 = 950 \text{ m}$. Time for A: $T_A = \frac{1000}{S_A}$; Time for B: $T_B = \frac{950}{S_B}$. A beats B by 10 seconds:

$$\frac{950}{S_B} - \frac{1000}{S_A} = 10 \quad \text{--- (Equation 1)}$$

• Scenario 2: B has a 20-second head start. When B finishes $1000 \text{ m}$, A is 50 m behind the finish line (covers $950 \text{ m}$). Time B ran: $\frac{1000}{S_B}$. Time A ran: $\frac{950}{S_A}$. Since B started 20 seconds earlier:

$$\frac{1000}{S_B} - \frac{950}{S_A} = 20 \quad \text{--- (Equation 2)}$$

• Step 2: Solve the linear system: Let $x = \frac{1}{S_B}$ and $y = \frac{1}{S_A}$:

$$950x - 1000y = 10 \implies 95x - 100y = 1 \quad \text{--- (1)}$$ $$1000x - 950y = 20 \implies 100x - 95y = 2 \quad \text{--- (2)}$$

Add (1) and (2):

$$195x - 195y = 3 \implies x - y = \frac{3}{195} = \frac{1}{65} \quad \text{--- (3)}$$

Subtract (1) from (2):

$$5x + 5y = 1 \implies x + y = \frac{1}{5} = \frac{13}{65} \quad \text{--- (4)}$$

Add (3) and (4):

$$2x = \frac{1}{65} + \frac{13}{65} = \frac{14}{65} \implies x = \frac{7}{65} \implies S_B = \frac{65}{7} \approx 9.29 \text{ m/s}$$

Subtract (3) from (4):

$$2y = \frac{13}{65} - \frac{1}{65} = \frac{12}{65} \implies y = \frac{6}{65} \implies S_A = \frac{65}{6} \approx 10.83 \text{ m/s}$$

• Conclusion: Runner A's speed is $\frac{65}{6}$ m/s ($\approx 10.83$ m/s), and Runner B's speed is $\frac{65}{7}$ m/s ($\approx 9.29$ m/s).

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Worked Example 3: Circular Testing Track Rendezvous Analysis

Problem: Two locomotives are undergoing continuous speed trials around a closed circular testing track of circumference $1200 \text{ meters}$. Locomotive $\alpha$ operates at a steady speed of $72 \text{ km/h}$, and Locomotive $\beta$ operates at $54 \text{ km/h}$. Both depart from the same start line at the same instant. Determine:

1. The time elapsed before they meet for the first time if they travel in opposite directions.
2. The time elapsed before they meet for the first time if they travel in the same direction.
3. The time elapsed before they meet for the first time at the exact starting line.
4. The number of distinct meeting points along the track in both cases.

• Step 1: Convert operational speeds to $\text{m/s}$:

$$S_\alpha = 72 \times \frac{5}{18} = 20 \text{ m/s}$$ $$S_\beta = 54 \times \frac{5}{18} = 15 \text{ m/s}$$

• Step 2: First meeting anywhere on track:
  • In opposite directions:

$$T_{\text{meet, opp}} = \frac{L}{S_\alpha + S_\beta} = \frac{1200}{20 + 15} = \frac{1200}{35} = \frac{240}{7} \approx 34.29 \text{ seconds}$$

• In the same direction:

$$T_{\text{meet, same}} = \frac{L}{S_\alpha - S_\beta} = \frac{1200}{20 - 15} = \frac{1200}{5} = 240 \text{ seconds} = 4 \text{ minutes}$$

• Step 3: First meeting at the starting point:
  • Lap time for $\alpha$: $t_\alpha = \frac{1200}{20} = 60 \text{ seconds}$
  • Lap time for $\beta$: $t_\beta = \frac{1200}{15} = 80 \text{ seconds}$

$$T_{\text{start}} = \text{LCM}(t_\alpha, t_\beta) = \text{LCM}(60, 80) = 240 \text{ seconds} = 4 \text{ minutes}$$

• Step 4: Distinct meeting points: Ratio of speeds in lowest co-prime terms:

$$\frac{S_\alpha}{S_\beta} = \frac{20}{15} = \frac{4}{3} \implies a = 4, b = 3, \quad \gcd(4, 3) = 1$$

• Same direction: Number of points $= a - b = 4 - 3 = 1$ (the starting point itself).
• Opposite directions: Number of points $= a + b = 4 + 3 = 7$ points (spaced every $\frac{1200}{7} \approx 171.43 \text{ m}$).
• Conclusion: Opposite meeting time is $\frac{240}{7}$ s, same direction meeting time is 240 s, start point meeting time is 240 s, with 1 point (same) and 7 points (opposite).

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