The Workbook

1. \(\vec v_{AB} = 3\hat i - 4\hat j\) m/s.

2. \(\sqrt{9 + 16} = 5\) m/s.

3. \(\tan\theta = \dfrac{4}{3}\), so \(\theta = 53^\circ\) south of east. Seen from \(B\), \(A\) moves east and falls behind southward.

\(\vec v_{AB} = 20\hat i - 20\hat j\): magnitude \(20\sqrt{2} \approx 28\) m/s, at \(45^\circ\) south of east.

\(\vec v_{\text{ground}} = \vec v_{\text{air}} + \vec v_{\text{wind}} = 80\hat j + 60\hat i\). Magnitude \(\sqrt{6400 + 3600} = 100\) m/s. \(\tan\phi = \dfrac{60}{80}\), so \(\phi = 37^\circ\) east of north: the wind pushes it off course.

The plane must point partly into the wind so that the westward part of its airspeed cancels the wind. \(100\sin\alpha = 60\), so \(\sin\alpha = 0.6\) and \(\alpha = 37^\circ\) west of north. Ground speed \(= 100\cos 37^\circ = 80\) m/s, all of it northward.

1. \(\vec v_{\text{rain,man}} = \vec v_{\text{rain}} - \vec v_{\text{man}} = -4\hat j - 3\hat i\).

2. \(\sqrt{9 + 16} = 5\) m/s. \(\tan\theta = \dfrac{3}{4}\), so \(\theta = 37^\circ\) from the vertical.

3. The rain seems to come from the front, so he tilts the umbrella forward by \(37^\circ\).

1. \(\tan 45^\circ = \dfrac{v_{\text{man}}}{v_{\text{rain}}} = 1\), so \(v_{\text{rain}} = 6\) m/s.

2. \(\sqrt{6^2 + 6^2} = 6\sqrt{2} \approx 8.5\) m/s.

3. \(\tan\phi = \dfrac{12}{6} = 2\), so \(\phi \approx 63^\circ\): faster walking makes the rain seem more slanted.

\(\tan 37^\circ = \dfrac{v_{\text{man}}}{v_{\text{rain}}}\), so \(\dfrac{3}{4} = \dfrac{3}{v_{\text{rain}}}\) and \(v_{\text{rain}} = 4\) m/s.

Rain velocity: horizontal \(10\sin 37^\circ = 6\) m/s towards the west, vertical \(8\) m/s down. For the rain to appear vertical, the man must move with its horizontal part: \(6\) m/s towards the west. Then \(\vec v_{\text{rain}} - \vec v_{\text{man}}\) has no horizontal component.

\(\vec v_{\text{rain}} = \vec v_{\text{rain,man}} + \vec v_{\text{man}} = -4\hat j + 3\hat i\). Speed \(= 5\) m/s, at \(\tan^{-1}\dfrac{3}{4} = 37^\circ\) from the vertical, leaning towards the east.

1. Only the boat's own velocity carries it across: \(t = \dfrac{100}{4} = 25\) s.

2. The river carries it \(2 \times 25 = 50\) m downstream.

3. \(\sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5} \approx 4.5\) m/s.

1. The upstream part of the boat's velocity must cancel the current: \(4\sin\theta = 2\), so \(\theta = 30^\circ\).

2. Ground speed \(= 4\cos 30^\circ = 2\sqrt{3} \approx 3.5\) m/s, straight across. Time \(= \dfrac{100}{2\sqrt{3}} = \dfrac{50}{\sqrt{3}} \approx 29\) s.

3. Question 10 is quicker (\(25\) s). Heading straight across uses the whole \(4\) m/s to cross; heading upstream spends part of it fighting the current.

Shortest time: head perpendicular, \(t_{\min} = \dfrac{d}{v_b}\), with drift \(v_r t_{\min} = \dfrac{v_r d}{v_b}\). Shortest path: head upstream with \(v_b\sin\theta = v_r\); the speed across is \(\sqrt{v_b^2 - v_r^2}\), so \(t = \dfrac{d}{\sqrt{v_b^2 - v_r^2}}\).

1. No. To cancel the current it would need \(3\sin\theta = 5\), which is impossible; the boat is slower than the river, so it is always carried downstream.

2. \(\cos\theta = \dfrac{3}{5}\), \(\sin\theta = \dfrac{4}{5}\). Speed across \(= 3 \times 0.8 = 2.4\) m/s, so \(t = \dfrac{100}{2.4}\) s. Downstream speed \(= 5 - 3 \times 0.6 = 3.2\) m/s. Drift \(= 3.2 \times \dfrac{100}{2.4} = \dfrac{400}{3} \approx 133\) m.

1. \(t = \dfrac{20}{1} = 20\) s; drift \(= 0.5 \times 20 = 10\) m.

2. \(\sin\theta = \dfrac{0.5}{1}\), \(\theta = 30^\circ\) upstream. Speed across \(= \cos 30^\circ = \dfrac{\sqrt{3}}{2}\) m/s, so \(t = \dfrac{40}{\sqrt{3}} \approx 23\) s.

\(t_1 = \dfrac{d}{v_b}\) and \(t_2 = \dfrac{d}{\sqrt{v_b^2 - v_r^2}}\). \(\dfrac{t_2}{t_1} = \dfrac{1}{\sqrt{1 - (v_r/v_b)^2}} = 1.25\), so \(1 - (v_r/v_b)^2 = 0.64\) and \(\dfrac{v_r}{v_b} = 0.6\).

To go straight across the swimmer would need \(2\sin\theta = 2\), that is, \(\theta = 90^\circ\), pointing straight upstream with no speed left to cross. So no. Shortest time: \(\dfrac{400}{2} = 200\) s, with drift \(2 \times 200 = 400\) m.

1. \(\vec v_{BA} = \vec v_B - \vec v_A = -3\hat i - 4\hat j\): magnitude \(5\) m/s, pointing \(53^\circ\) south of west.

2. From \(A\), \(B\) starts at \(\vec r_0 = 100\hat i\) and moves along \(\vec v_{BA}\). The least distance is the perpendicular from \(A\) to that line: \(d = r_0\sin\alpha\), where \(\alpha\) is the angle between \(\vec r_0\) and \(\vec v_{BA}\). \(\sin\alpha = \dfrac{4}{5}\), so \(d = 80\) m.

3. \(B\) must travel \(r_0\cos\alpha = 60\) m along the line at \(5\) m/s: \(t = 12\) s.

1. \(\vec v_{BA} = -20\hat j - 20\hat i\): \(20\sqrt{2}\) m/s towards the south-west.

2. \(\vec r_0 = 100\hat j\) makes \(45^\circ\) with \(\vec v_{BA}\): \(d = 100\sin 45^\circ = 50\sqrt{2} \approx 71\) m.

3. \(t = \dfrac{r_0\cos 45^\circ}{v} = \dfrac{100/\sqrt{2}}{20\sqrt{2}} = 2.5\) s.

\(P\) reaches the crossing after \(\dfrac{200}{20} = 10\) s and \(Q\) after \(\dfrac{150}{15} = 10\) s. They arrive together, so they collide. In relative terms the relative velocity points exactly along the initial separation, so the least distance is zero.

\(\vec r_0 = 100\hat i\), \(\vec v_{BA} = 5\hat j - 5\hat i\), of magnitude \(5\sqrt{2}\) at \(135^\circ\) to \(\vec r_0\). \(d = 100\sin 135^\circ = 50\sqrt{2} \approx 71\) m. Time \(= \dfrac{-\vec r_0\cdot\vec v}{v^2} = \dfrac{500}{50} = 10\) s.

Seen from \(A\), \(B\) moves along a straight line through \(\vec r_0\) in the direction of \(\vec v\). The least distance from \(A\) to this line is \(r_0\sin\alpha = \dfrac{|\vec r_0\times\vec v|}{|\vec v|}\).

\(B\) reaches the foot of the perpendicular after moving \(-r_0\cos\alpha\) along the line (the minus sign because \(\vec v\) points back towards \(A\)): \(t = \dfrac{-\vec r_0\cdot\vec v}{|\vec v|^2}\). Question 20: \(|\vec r_0\times\vec v| = 100 \times 5 = 500\), \(|\vec v| = 5\sqrt{2}\), so \(d = \dfrac{500}{5\sqrt{2}} \approx 71\) m and \(t = \dfrac{500}{50} = 10\) s.

A flag points along the wind relative to the car: \(\vec v_{\text{wind}} - \vec v_{\text{car}} = 20\hat i - 20\hat j\), towards the south-east. The car's own motion adds an apparent wind from the north.

Ground velocity \(= 4\hat j + 3\hat i\): speed \(5\) m/s, at \(\tan^{-1}\dfrac{4}{3} = 53^\circ\) to the bank, leaning downstream.

Across the river the boat's velocity component is \(v_b\sin\theta\), so \(t = \dfrac{d}{v_b\sin\theta}\). Along the river the ground velocity is \(v_r - v_b\cos\theta\), so the drift is \((v_r - v_b\cos\theta)t\).

\(\theta = 90^\circ\): \(t = \dfrac{d}{v_b}\), the shortest time, with drift \(\dfrac{v_r d}{v_b}\).