Question

If the riverboat is crossing the river, with a velocity of 4 m/s along the north and the river is flowing with a velocity of 3 m/s along the west,


then what is the magnitude and direction of resultant velocity?

• A. 5 m/s in south-east direction
• B. 7 m/s in north-west direction
• C. 5 m/s in north-west direction
• D. 7 m/s in south-east direction

Answer 1

The correct option is C 5 m/s in north-west direction

We know that the magnitude of resultant can be calculated by

​​​​$R=\sqrt{{V_{rb}}^2+{V_r}^2+2V_{rb}\times V_r\cos \theta }$
(By triangular law of vector addition)

where, $V_{rb}=$ velocity of riverboat
and $V_r=$ velocity of river

but here $\theta ={90}^{\circ }$
So,
$R=\sqrt{3^2+4^2}$
$\Rightarrow R=5m/s$
For the direction, we can calculate

$\alpha ={\tan }^{-1}\left(\frac{V_r}{V_{rb}}\right)={\tan }^{-1}\left(\frac{3}{4}\right)=36.{87}^o$

So, here we get the direction of resultant velocity as 36.87° in north-west direction.
So, option C is correct.