Source: Principles of Physics
Problem Sets:
Problem
Ship A is located 4.0 km north and 2.5 km east of ship B. Ship A has a velocity of 22 km/h toward the south, and ship B has a velocity of 40 km/h in a direction 37° north of east. Let $t=0$ when the ships are in the positions described.
- What is the velocity of A relative to B in unit-vector notation with $\hat{i}$ toward the east?
- Write an expression (in terms of $\hat{i}$ and $\hat{j}$) for the position of A relative to B as a function of $t$.
- At what time is the separation between the ships least?
- What is that least separation?
[Q1] The velocity of A relative to B is $\vec{v}_{AB} = (-32\hat{i} - 46\hat{j})$ km/h. [Q2] The position of A relative to B is $\vec{r}_{AB}(t) = (2.5 - 32t)\hat{i} + (4.0 - 46t)\hat{j}$ km. [Q3] The time of least separation is $t_{min} = -\frac{\vec{r}_{AB0} \cdot \vec{v}_{AB}}{|\vec{v}_{AB}|^2} \approx 0.084$ h. [Q4] The least separation is $d_{min} = |\vec{r}_{AB}(t_{min})| \approx 0.22$ km.
We define a coordinate system with the origin at the initial position of ship B, $\hat{i}$ pointing east, and $\hat{j}$ pointing north.
The initial positions and velocities of the ships are:
- Ship A: $\vec{r}_{A0} = (2.5\hat{i} + 4.0\hat{j})$ km, $\vec{v}_A = -22\hat{j}$ km/h
- Ship B: $\vec{r}_{B0} = 0$, $\vec{v}_B = (40\cos37^\circ)\hat{i} + (40\sin37^\circ)\hat{j}$ km/h
[Q1] Velocity of A relative to B
The velocity of ship A relative to ship B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.
First, calculate the components of $\vec{v}_B$:
$\vec{v}_B \approx (40(0.799))\hat{i} + (40(0.602))\hat{j} \approx (32.0\hat{i} + 24.1\hat{j})$ km/hNow, subtract $\vec{v}_B$ from $\vec{v}_A$:
$$\vec{v}_{AB} = (-22\hat{j}) - (32.0\hat{i} + 24.1\hat{j})$$ $$\vec{v}_{AB} = -32.0\hat{i} - (22 + 24.1)\hat{j}$$ $$\vec{v}_{AB} = (-32.0\hat{i} - 46.1\hat{j}) \text{ km/h}$$[Q2] Position of A relative to B
The position of A relative to B as a function of time $t$ is $\vec{r}_{AB}(t) = \vec{r}_{AB0} + \vec{v}_{AB}t$. The initial relative position is $\vec{r}_{AB0} = \vec{r}_{A0} - \vec{r}_{B0} = (2.5\hat{i} + 4.0\hat{j})$ km.
Substituting the vectors:
$$\vec{r}_{AB}(t) = (2.5\hat{i} + 4.0\hat{j}) + (-32.0\hat{i} - 46.1\hat{j})t$$ $$\vec{r}_{AB}(t) = (2.5 - 32.0t)\hat{i} + (4.0 - 46.1t)\hat{j} \text{ km}$$[Q3] Time of Least Separation
The separation is least when its square, $d^2 = |\vec{r}_{AB}(t)|^2$, is at a minimum. In the reference frame of ship B, ship A moves along a straight line with velocity $\vec{v}_{AB}$. The minimum distance from B (the origin) to this line occurs when the position vector $\vec{r}_{AB}(t)$ is perpendicular to the velocity vector $\vec{v}_{AB}$.
The condition for minimum separation is $\vec{r}_{AB}(t) \cdot \vec{v}_{AB} = 0$.
$$(\vec{r}_{AB0} + \vec{v}_{AB}t) \cdot \vec{v}_{AB} = 0$$ $$\vec{r}_{AB0} \cdot \vec{v}_{AB} + t(\vec{v}_{AB} \cdot \vec{v}_{AB}) = 0$$Solving for the time of minimum separation, $t_{min}$:
$$t_{min} = - \frac{\vec{r}_{AB0} \cdot \vec{v}_{AB}}{|\vec{v}_{AB}|^2}$$We calculate the dot product and the squared magnitude using unrounded values for accuracy:
$\vec{v}_{AB} \approx (-31.95\hat{i} - 46.07\hat{j})$ km/h $\vec{r}_{AB0} \cdot \vec{v}_{AB} = (2.5)(-31.95) + (4.0)(-46.07) = -79.875 - 184.28 = -264.16$ km²/h $|\vec{v}_{AB}|^2 = (-31.95)^2 + (-46.07)^2 = 1020.8 + 2122.4 = 3143.2$ (km/h)² $$t_{min} = - \frac{-264.16}{3143.2} \approx 0.08404 \text{ h}$$[Q4] Least Separation
The least separation $d_{min}$ is the magnitude of the relative position vector at $t_{min}$.
$$d_{min} = |\vec{r}_{AB}(t_{min})|$$We substitute $t_{min}$ into the expression for $\vec{r}_{AB}(t)$:
$x(t_{min}) = 2.5 - 31.95(0.08404) \approx 2.5 - 2.685 \approx -0.185$ km $y(t_{min}) = 4.0 - 46.07(0.08404) \approx 4.0 - 3.872 \approx 0.128$ km $$d_{min} = \sqrt{(-0.185)^2 + (0.128)^2} = \sqrt{0.0342 + 0.0164} = \sqrt{0.0506} \approx 0.225 \text{ km}$$