$\begin{array}{1 1}\text{11.2m/s at 26.6}^{\large\circ}\text{ E of N}\\\text{8.66m/s at 63.4}^{\large\circ}\text{ W of N}\\\text{11.2m/s at 63.4}^{\large\circ}\text{ W of N}\\\text{8.66m/s at 26.6}^{\large\circ}\text{ E of N}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : $\text{11.2m/s at 63.4}^{\large\circ}\text{ W of N}$

The river is flowing at 10m/s to the right(east) and the resultant desired velocity is 5m/s up (north).

Therefore,the actual velocity, relative to the river ,is heading W of N.

$\Rightarrow$ Using Pythagoras theorem, we can calculate the velocity of the boat relative to the water as $v_B = \sqrt { 10^2 + 5^2} = 11.2\;m/s$

The angle is given by the tangent function:

From the diagram above, $\tan\theta=\large\frac{10}{5}$$=2$.Therefore $\theta=63.4^{\large\circ}$ W of N.

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...