Entrance Exams Q&A: Q. 1.6 -- "Problems in General Physics" by I.E. Irodov

June 9, 2008

Q. 1.6 -- "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$ :
A ship moves along the equator to the east with velocity = 30km/hour. The southeastern wind bloes at an angle $\phi$ = to the equator with velocity = 15km/hour. Find the wind velocity relative to the ship and the angle $\phi'$ between the equator and the wind direction in the reference frame fixed to the ship.

$\textbf{Answer}$ :
Image diag

Horizontal component of wind velocity w.r.t ship = $v - v_0 \cos \phi$

Vertical component of wind velocity w.r.t ship = $v_0 \sin \phi$

Wind speed relative to ship = $\sqrt {(v - (-v_0 \cos \phi))^2 + (v_0 \sin \phi)^2}$

= $\sqrt {v^2 + v_0^2 + 2v v_0 \cos \phi}$

Angle relative to ship = $v_0sin\phi/(v+v_0\cos\phi)$

Using values:
Relative speed = $\sqrt{900 + 225 + 2 \times 30 \times 15 cos 60^o}$ = $\sqrt{1575}$ $\approx$ 39.6 km/hour
Wind angle = $tan^{-1}(15 \times \sin 60^o)/(30 + 15\cos 60^o)$ $\approx 19^o$

2 comments:

Addu5221 said...: I have a Doubt. Ithink It should be sqrt(v1^2 + v2^2 -2v1v2cosB); April 26, 2015 at 1:16 PM
Apurva Sharan said...: Hi Adnan,

I guess you are getting confused about the wind direction. wind direction is denoted by the direction FROM which wind is blowing (and not TO). Refer to the diagram -- that's the correct representation of wind direction.

https://www.google.co.in/search?q=What+does+wind+direction+mean

Regards,
Apurva; August 4, 2015 at 10:06 PM

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