Solution to Question 30
1. Problem Analysis
The boatman always steers the boat towards the fixed target point directly opposite the starting point.
- River width $b = 320 \text{ m}$.
- Boat speed relative to water $v = v_{b/w} = 2.5 \text{ m/s}$.
- River current speed $u = v_w = 1.5 \text{ m/s}$.
Since $v > u$, the boat will eventually reach the target point. This is a “Curve of Pursuit” problem.
2. Decomposition of Velocities
Let the instantaneous position of the boat be at a distance $r$ from the target $B$. Let the line of sight make an angle $\theta$ with the vertical ($y$-axis).
The velocity components are:
-
Along the line of sight (Radial direction):
The boat moves towards $B$ with speed $v$, while the current component $u \sin \theta$ moves it away.
$$ v_r = \frac{dr}{dt} = -v + u \sin \theta \quad \dots(1) $$
-
Horizontal direction (Along river flow):
The boat moves upstream with component $v \sin \theta$ and downstream with current $u$.
$$ v_x = \frac{dx}{dt} = u – v \sin \theta \quad \dots(2) $$
3. Deriving the Time Equation
We can eliminate the $\sin \theta$ term by a linear combination of the two equations. Multiply equation (1) by $v$ and equation (2) by $u$:
$$ v \frac{dr}{dt} = -v^2 + uv \sin \theta $$
$$ u \frac{dx}{dt} = u^2 – uv \sin \theta $$
Adding these two equations:
$$ v \frac{dr}{dt} + u \frac{dx}{dt} = -(v^2 – u^2) $$
Integrating with respect to time from start ($t=0$) to finish ($t=\tau$):
$$ \int_{start}^{end} \left( v \, dr + u \, dx \right) = \int_{0}^{\tau} -(v^2 – u^2) \, dt $$
4. Boundary Conditions and Calculation
Limits of integration:
- Radial distance $r$: Changes from initial separation $b$ to $0$ (since the boat reaches the target). $\Delta r = 0 – b = -b$.
- Horizontal position $x$: The boat starts at $x=0$ and, since it successfully lands at the target directly opposite, the final $x=0$. $\Delta x = 0 – 0 = 0$.
Substituting these into the integrated equation:
$$ v(-b) + u(0) = -(v^2 – u^2) \tau $$
$$ -vb = -(v^2 – u^2) \tau $$
$$ \tau = \frac{vb}{v^2 – u^2} $$
Numerical Substitution:
$$ \tau = \frac{2.5 \times 320}{(2.5)^2 – (1.5)^2} $$
$$ \tau = \frac{800}{6.25 – 2.25} = \frac{800}{4.0} $$
$$ \tau = 200 \text{ s} $$
Answer: The time taken to cross the river is 200 s.
1. Problem Analysis
The boatman always steers the boat towards the fixed target point directly opposite the starting point.
- River width $b = 320 \text{ m}$.
- Boat speed relative to water $v = v_{b/w} = 2.5 \text{ m/s}$.
- River current speed $u = v_w = 1.5 \text{ m/s}$.
2. Decomposition of Velocities
Let the instantaneous position of the boat be at a distance $r$ from the target $B$. Let the line of sight make an angle $\theta$ with the vertical ($y$-axis). The velocity components are:
- Along the line of sight (Radial direction): The boat moves towards $B$ with speed $v$, while the current component $u \sin \theta$ moves it away. $$ v_r = \frac{dr}{dt} = -v + u \sin \theta \quad \dots(1) $$
- Horizontal direction (Along river flow): The boat moves upstream with component $v \sin \theta$ and downstream with current $u$. $$ v_x = \frac{dx}{dt} = u – v \sin \theta \quad \dots(2) $$
3. Deriving the Time Equation
We can eliminate the $\sin \theta$ term by a linear combination of the two equations. Multiply equation (1) by $v$ and equation (2) by $u$: $$ v \frac{dr}{dt} = -v^2 + uv \sin \theta $$ $$ u \frac{dx}{dt} = u^2 – uv \sin \theta $$ Adding these two equations: $$ v \frac{dr}{dt} + u \frac{dx}{dt} = -(v^2 – u^2) $$ Integrating with respect to time from start ($t=0$) to finish ($t=\tau$): $$ \int_{start}^{end} \left( v \, dr + u \, dx \right) = \int_{0}^{\tau} -(v^2 – u^2) \, dt $$
4. Boundary Conditions and Calculation
Limits of integration:
- Radial distance $r$: Changes from initial separation $b$ to $0$ (since the boat reaches the target). $\Delta r = 0 – b = -b$.
- Horizontal position $x$: The boat starts at $x=0$ and, since it successfully lands at the target directly opposite, the final $x=0$. $\Delta x = 0 – 0 = 0$.
Numerical Substitution: $$ \tau = \frac{2.5 \times 320}{(2.5)^2 – (1.5)^2} $$ $$ \tau = \frac{800}{6.25 – 2.25} = \frac{800}{4.0} $$ $$ \tau = 200 \text{ s} $$