Question

A motor cylist is riding north in still air at $36km{h}^{-1}$. the wind starts blowing westward with a velocity $18km{h}^{-1}$ . the direction of apparent velocity is

• A. $ta{n}^{-1}\left(\frac{1}{2}\right)$ west of north
• B. $ta{n}^{-1}\left(\frac{1}{2}\right)$ north of west
• C. $ta{n}^{-1}\left(2\right)$ east of north
• D. $ta{n}^{-1}\left(2\right)$ north of east

Answer 1

Answer: (A)

$ta{n}^{-1}\left(\frac{1}{2}\right)$ west of north

Let the velocity of the motor cyclist is $V_m$;

The velocity off the wind is $V_w$;

Given: $V_m=36kmh{r}^{-1}$ and $V_w=18kmh{r}^{-1}$;

Convert the above given velocities in m\s;

$V_m=36kmh{r}^{-1}=\frac{36\times 5}{18}m{s}^{-1}=10m{s}^{-1}$;

$V_w=18kmh{r}^{-1}=\frac{18\times 5}{18}m{s}^{-1}$;

From the diagram we can show that:

$tan\theta =\frac{5}{10}$

$tan\theta =\frac{1}{2}$

$\theta =ta{n}^{-1}\left(\frac{1}{2}\right)$

Therefore, the direction of the apparent velocity is: $\theta =ta{n}^{-1}\left(\frac{1}{2}\right)$