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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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Two persons P and Q are standing 54m apart on a long horizontal belt moving with a speed of $4ms^{-1}$ in the direction from P to Q.Person P rolls a round stone towards person Q with speed of $9ms^{-1}$ with respect to the belt.The velocity of stone with respect to an observer on a stationary platform is

$\begin{array}{1 1}13ms^{-1}\text{ in the direction to motion of the belt}\\13ms^{-1}\text{ opposite to the direction of motion of the belt}\\5ms^{-1}\text{ in the direction to motion of the belt}\\5ms^{-1}\text{ opposite to the direction of motion of the belt}\end{array} $

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Answer : $13ms^{-1}\text{ in the direction to motion of the belt}$
Let us choose the positive direction to be the direction from P to Q.i.e., the direction of motion of the belt.Then,the velocity of the belt is
$v_B=+4ms^{-1}$
Speed of stone with respect to the belt is
$v_{SB}=+9ms^{-1}$
If $v_S$ is the speed of the stone with respect to a stationary observer,we have
$v_{SB}=v_S-v_B$
or $v_S=v_{SB}+v_B=+9+4=+13ms^{-1}$
The positive sign shows that the direction of velocity of stone is from P to Q,i.e in the direction of motion of the belt.Hence the correct choice is $13ms^{-1}\text{ in the direction to motion of the belt}$
answered Aug 14, 2014 by sreemathi.v
 

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