$\begin{array}{1 1}11.2 km/h, 26.6^{\large\circ}\\12.2 km/h, 36.6^{\large\circ}\\13.2 km/h, 25.6^{\large\circ}\\10.2 km/h, 16.6^{\large\circ}\end{array} $

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Answer : 11.2 km/h, 26.6$^{\large\circ}$

Let $V_{br}$,the velocity of the boat relative to the river,and $V_{rE}$,the velocity of the river relative to earth.What we must find is $V_{bE}$,the velocity of the boat relative to Earth.The relationship between these three quantities is

$V_{bE}=V_{br}+V_{rE}$

The quantity $V_{br}$ is due north,$V_{rE}$ is due east,and the vector sum of the two,$V_{bE}$ is at an angle $\theta$

We can find the speed $v_{bE}$ of the boat relative to Earth by using the Pythagorean theorem :

$v_{bE}=\sqrt{v_{br}^2+v_{rE}^2}=\sqrt{(10.0)^2+(5.00)^2}$ km/h

$\Rightarrow 11.2 km/h$

The direction of $v_{bE}$ is

$\theta=\tan^{-1}\big(\large\frac{v_{rE}}{v_{br}}\big)$$=\tan^{-1}\big(\large\frac{5.00}{10.0}\big)$$=26.6^{\large\circ}$

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