Solution to Question 31
1. Relative Velocity Setup
Let us analyze the motion of Ship B relative to Ship A. We treat Ship A as stationary at the origin $(0,0)$.
- Velocity of A ($\vec{v}_A$): $16\sqrt{2}$ km/h towards Northeast ($45^\circ$).
$$ \vec{v}_A = 16\sqrt{2}(\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = 16\hat{i} + 16\hat{j} \text{ km/h} $$
- Velocity of B ($\vec{v}_B$): 4 km/h towards North.
$$ \vec{v}_B = 4\hat{j} \text{ km/h} $$
- Relative Velocity of B w.r.t A ($\vec{v}_{B/A}$):
$$ \vec{v}_{B/A} = \vec{v}_B – \vec{v}_A = 4\hat{j} – (16\hat{i} + 16\hat{j}) = -16\hat{i} – 12\hat{j} \text{ km/h} $$
Magnitude: $|\vec{v}_{B/A}| = \sqrt{(-16)^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \text{ km/h}$.
2. Relative Position and Path
At midnight ($t=0$), Ship B is 80 km East of Ship A.
- Initial Position of B relative to A: $\vec{r}_{B/A} = 80\hat{i} \text{ km}$.
- Communication Range: $R = 50 \text{ km}$.
In this frame, Ship B moves along a straight line starting from $(80, 0)$ with velocity vector $(-16, -12)$. Communication is possible when B is inside the circle of radius $R=50$ centered at A.
3. Calculation of Time Interval
Distance of closest approach ($d$):
The equation of the line passing through $(80,0)$ with slope $m = \frac{-12}{-16} = \frac{3}{4}$ is:
$$ y – 0 = \frac{3}{4}(x – 80) \implies 3x – 4y – 240 = 0 $$
The perpendicular distance from origin $A(0,0)$ to this line is:
$$ d = \frac{|3(0) – 4(0) – 240|}{\sqrt{3^2 + (-4)^2}} = \frac{240}{5} = 48 \text{ km} $$
Since $d = 48 < 50$, the ship enters the communication range. The length of the path inside the circle (chord length) is $2l$, where:
$$ l = \sqrt{R^2 - d^2} = \sqrt{50^2 - 48^2} = \sqrt{2500 - 2304} = \sqrt{196} = 14 \text{ km} $$
Total distance covered while in communication:
$$ D_{comm} = 2l = 2 \times 14 = 28 \text{ km} $$
Time Interval:
$$ \Delta t = \frac{D_{comm}}{|\vec{v}_{B/A}|} = \frac{28}{20} = 1.4 \text{ hours} $$
Converting to minutes: $0.4 \times 60 = 24$ minutes.
So, $\Delta t = 1 \text{ hour } 24 \text{ minutes}$.
4. Specific Timestamps
Distance from Start ($80,0$) to the midpoint of the chord (foot of perpendicular) is:
$$ D_{mid} = \sqrt{80^2 – 48^2} = 64 \text{ km} $$
Time to enter range ($t_1$):
$$ t_1 = \frac{D_{mid} – l}{v_{rel}} = \frac{64 – 14}{20} = \frac{50}{20} = 2.5 \text{ hours} $$
$2.5 \text{ hours} = 2:30 \text{ AM}$.
Time to leave range ($t_2$):
$$ t_2 = \frac{D_{mid} + l}{v_{rel}} = \frac{64 + 14}{20} = \frac{78}{20} = 3.9 \text{ hours} $$
$3.9 \text{ hours} = 3 \text{ hours } 54 \text{ minutes} \implies 3:54 \text{ AM}$.
Answer: The ships are in communication from 2:30 A.M. to 3:54 A.M.
1. Relative Velocity Setup
Let us analyze the motion of Ship B relative to Ship A. We treat Ship A as stationary at the origin $(0,0)$.
- Velocity of A ($\vec{v}_A$): $16\sqrt{2}$ km/h towards Northeast ($45^\circ$). $$ \vec{v}_A = 16\sqrt{2}(\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = 16\hat{i} + 16\hat{j} \text{ km/h} $$
- Velocity of B ($\vec{v}_B$): 4 km/h towards North. $$ \vec{v}_B = 4\hat{j} \text{ km/h} $$
- Relative Velocity of B w.r.t A ($\vec{v}_{B/A}$): $$ \vec{v}_{B/A} = \vec{v}_B – \vec{v}_A = 4\hat{j} – (16\hat{i} + 16\hat{j}) = -16\hat{i} – 12\hat{j} \text{ km/h} $$ Magnitude: $|\vec{v}_{B/A}| = \sqrt{(-16)^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \text{ km/h}$.
2. Relative Position and Path
At midnight ($t=0$), Ship B is 80 km East of Ship A.
- Initial Position of B relative to A: $\vec{r}_{B/A} = 80\hat{i} \text{ km}$.
- Communication Range: $R = 50 \text{ km}$.
3. Calculation of Time Interval
Distance of closest approach ($d$): The equation of the line passing through $(80,0)$ with slope $m = \frac{-12}{-16} = \frac{3}{4}$ is: $$ y – 0 = \frac{3}{4}(x – 80) \implies 3x – 4y – 240 = 0 $$ The perpendicular distance from origin $A(0,0)$ to this line is: $$ d = \frac{|3(0) – 4(0) – 240|}{\sqrt{3^2 + (-4)^2}} = \frac{240}{5} = 48 \text{ km} $$
Since $d = 48 < 50$, the ship enters the communication range. The length of the path inside the circle (chord length) is $2l$, where: $$ l = \sqrt{R^2 - d^2} = \sqrt{50^2 - 48^2} = \sqrt{2500 - 2304} = \sqrt{196} = 14 \text{ km} $$ Total distance covered while in communication: $$ D_{comm} = 2l = 2 \times 14 = 28 \text{ km} $$
Time Interval: $$ \Delta t = \frac{D_{comm}}{|\vec{v}_{B/A}|} = \frac{28}{20} = 1.4 \text{ hours} $$ Converting to minutes: $0.4 \times 60 = 24$ minutes. So, $\Delta t = 1 \text{ hour } 24 \text{ minutes}$.
4. Specific Timestamps
Distance from Start ($80,0$) to the midpoint of the chord (foot of perpendicular) is:
$$ D_{mid} = \sqrt{80^2 – 48^2} = 64 \text{ km} $$
Time to enter range ($t_1$):
$$ t_1 = \frac{D_{mid} – l}{v_{rel}} = \frac{64 – 14}{20} = \frac{50}{20} = 2.5 \text{ hours} $$
$2.5 \text{ hours} = 2:30 \text{ AM}$.
Time to leave range ($t_2$):
$$ t_2 = \frac{D_{mid} + l}{v_{rel}} = \frac{64 + 14}{20} = \frac{78}{20} = 3.9 \text{ hours} $$
$3.9 \text{ hours} = 3 \text{ hours } 54 \text{ minutes} \implies 3:54 \text{ AM}$.