Question

On a day when the wind is blowing toward the south at 3 m/s, a runner jogs west at 4 m/s. What is the speed of the
air relative to the runner?

Answer 1

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

• ${\sf {V}_{w} }$= velocity of wind = 3 m/s (south)

• ${\sf {V}_{r} }$= velocity of runner = 4 m/s (west)

So,

• ${\sf {V}_{w} }$ = - 3 ${ \sf \hat{j}}$

[ taking unit vector along north be ${ \sf \hat{j}}$ ]

• ${\sf {V}_{r} }$ = - 4 ${ \sf \hat{i}}$

[ taking unit vector along west be ${ \sf \hat{i}}$ ]

$\qquad \dashrightarrow\sf \: V_{wr}= V_{w} - V_{r }$

[ where ${ \sf V_{wr}= }$ Relative velocity of wind with respect to runner ]

$\qquad \dashrightarrow\sf \: V_{wr}= - 3 \hat{ j} - ( - 4 \hat{i})$

$\qquad \dashrightarrow\sf \: V_{wr}= - 3 \hat{ j} + 4 \hat{i}$

That is : 3 m/s towards south and 4 m/s towards east.

So, it makes an angle of 37° with south, and 53° with East.

[ The resultant is depicted by yellow arrow in attachment ]

Magnitude :

$\qquad \dashrightarrow\sf \: |V_{wr} | = \sqrt{3 {}^{2} + 4 {}^{2} }$

$\qquad \dashrightarrow\sf \: |V_{wr} | = \sqrt{9 + 16 }$

$\qquad \dashrightarrow\sf \: |V_{wr} | = \sqrt{25 }$

$\qquad \dashrightarrow\sf \: |V_{wr} | = 5 \: \: m/s$

Direction :

$\dashrightarrow \sf \tan( \alpha) = \dfrac{b\: \sin \theta}{b + a \sin( \theta) }$

${\sf \theta }$ = angle between the two vectors ( along east and south) i.e 90°

• b = 4 units

$\dashrightarrow \sf \tan( \alpha) = \dfrac{4\: \sin (90 \degree)}{3+ 4 \cos( 90 \degree) }$

$\dashrightarrow \sf \tan( \alpha) = \dfrac{4 \times 1}{3+( 4 \times 0)}$

$\dashrightarrow \sf \tan( \alpha) = \dfrac{4}{3}$

$\dashrightarrow \sf \alpha = \tan {}^{ - 1} \bigg( \dfrac{4}{3 } \bigg)$

$\dashrightarrow \sf \alpha = 53 \degree$

So, it's direction is 53° east from south ~