Question

To a man walking at 7 kmph due west, the wind appears to blow from the north-west but when he walks at 3kmph due west, the wind appears to blow from north. The magnitude and actual direction of wind are

• A. $5kmph,ta{n}^{-1}\left(\frac{3}{4}\right)$ east of north
• B. $5kmph,ta{n}^{-1}\left(\frac{3}{4}\right)$ north of east
• C. $4kmph,ta{n}^{-1}\left(\frac{3}{4}\right)$ east of north
• D. $4kmph,ta{n}^{-1}\left(\frac{3}{4}\right)$ north of east

Answer 1

The correct option is C $4kmph,ta{n}^{-1}\left(\frac{3}{4}\right)$ east of north

let $v_m$and$v_a$ be the velocity of man and air respectively

let $v_a=v_1\overrightarrow{i}+v_2\overrightarrow{j}$

case 1 : $v_m=-7\overrightarrow{i}$

${\overrightarrow{v}}_{am}=(v_1+7)\overrightarrow{i}+v_2\overrightarrow{j}$

since ${\overrightarrow{v}}_{am}$ is along south-east (i.e. from north-west)

so $v_1+7=-v_2\dots \left(1\right)$

case 2:$v_m=-3\overrightarrow{i}$

${\overrightarrow{v}}_{am}=(v_1+3)\overrightarrow{i}+v_2\overrightarrow{j}$

since ${\overrightarrow{v}}_{am}$ is along south

$v_1+3=0\dots \left(2\right)$

using $\left(1\right)\&\left(2\right)$

$v_a=-3\overrightarrow{i}-4\overrightarrow{j}$

therefore wind blow at $5kmph$from $ta{n}^{-1}\left(\frac{3}{4}\right)$ east of north