logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
0 votes

A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The water in the river has a uniform speed of 5.00 km/h due east relative to the Earth. Determine the velocity of the boat relative to an observer standing on either bank.

$\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} $

Can you answer this question?
 
 

1 Answer

0 votes
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}$
answered Aug 14, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...