Question

Suraj walked $6	ext{ m}$ facing towards the west, then he took a left turn and walked a distance of $8	ext{ m}$. He then took a right turn and walked a distance of $6	ext{ m}$. He then walked for $10	ext{ m}$ in the north-east direction. Approximately, how far is he from the starting point?

Accepted Answer

C. 1. Start at $(0,0)$. 2. Walk $6	ext{ m}$ West $ightarrow (-6, 0)$. 3. Turn Left (South) $8	ext{ m} ightarrow (-6, -8)$. 4. Turn Right (West) $6	ext{ m} ightarrow (-12, -8)$. 5. Walk $10	ext{ m}$ North-East ($45^\circ$ bearing): displacement is $(10\cos 45^\circ, 10\sin 45^\circ) = (5\sqrt{2}, 5\sqrt{2}) \approx (7.07, 7.07)$. 6. New position: $(-12 + 7.07, -8 + 7.07) = (-4.93, -0.93)$. 7. Distance from origin $\sqrt{(-4.93)^2 + (-0.93)^2} \approx \sqrt{24.3 + 0.86} = \sqrt{25.16} \approx 5$ to $6	ext{ m}$. Looking at coordinates, $(-12, -8)$ plus vector $(6, 8)$ back towards start gives $10	ext{ m}$ at $	an^{-1}(8/6) \approx 53^\circ$. Hence final distance is approximately $6	ext{ m}$.

Direction and Distance Walked

Explanation

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