
A man is walking on a road with a velocity 3km/hr. Suddenly the rain starts falling. The velocity of rain is 10km/hr in a vertically downward direction. the relative velocity of rain with respect to man is:-
A. $\sqrt {13} \;{\rm{km/h}}$
B. $\sqrt 7 \;{\rm{km/h}}$
C. $\sqrt {109} \;{\rm{km/h}}$
D. $13\;{\rm{km/h}}$
Answer
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Hint:We know that the relative velocity is the net velocity of one body concerning another body. When the first body and second body both are in motion, then the relative rate is the difference between the velocities of first and second.
Complete step by step answer:
Since the direction of the rain is in negative y-direction so the velocity of the rain can be represented as,
${V_R} = - 10\hat j\;{\rm{km/h}}$
Since the direction of the man is in the positive X direction so the velocity of the man can be represented as,
${V_m} = 3\hat i\;{\rm{km/h}}$
The relative of rain with respect to the man ${V_{RM}}$ is,
${V_{RM}} = {V_R} - {V_M}$…….(I)
We will now substitute ${V_R} = - 10\hat j\;{\rm{km/h}}$, ${V_m} = 3\hat i\;{\rm{km/h}}$ in equation (I).
$\begin{array}{c}
{V_{RM}} = - 10\hat j\;{\rm{km/h}} - 3\hat i\;{\rm{km/h}}\\
= - 10\hat j\; - 3\hat i
\end{array}$
Therefore, the magnitude of ${V_{RM}}$ is,
\[\begin{array}{c}
\left| {{V_{RM}}} \right| = \sqrt {{{\left( { - 10} \right)}^2} + {3^2}} \\
= \sqrt {109} \;{\rm{km/h}}
\end{array}\]
Therefore, the velocity of the rain with respect to man is \[\sqrt {109} \;{\rm{km/h}}\] , and the correct option is (C).
Additional information:
We know that after the resultant velocity is calculated, the direction of the velocity is also of utmost importance. The line of the resultant of two vectors is close to the vector with large magnitude, and the direction of the resultant velocities and vectors are generally calculated with the positive direction of x-axis in radians or degree. If the angle is taken in an anti-clockwise order from the positive x-axis, then the value of the angle is positive and vice versa.
Note: We must follow a proper sign convention for determining the relative velocities and taking the directions appropriately. The velocity in the positive x-direction is taken as positive, and the velocity in the positive y-direction is taken as positive. Rest in all the other options the directions are negative.
Complete step by step answer:
Since the direction of the rain is in negative y-direction so the velocity of the rain can be represented as,
${V_R} = - 10\hat j\;{\rm{km/h}}$
Since the direction of the man is in the positive X direction so the velocity of the man can be represented as,
${V_m} = 3\hat i\;{\rm{km/h}}$
The relative of rain with respect to the man ${V_{RM}}$ is,
${V_{RM}} = {V_R} - {V_M}$…….(I)
We will now substitute ${V_R} = - 10\hat j\;{\rm{km/h}}$, ${V_m} = 3\hat i\;{\rm{km/h}}$ in equation (I).
$\begin{array}{c}
{V_{RM}} = - 10\hat j\;{\rm{km/h}} - 3\hat i\;{\rm{km/h}}\\
= - 10\hat j\; - 3\hat i
\end{array}$
Therefore, the magnitude of ${V_{RM}}$ is,
\[\begin{array}{c}
\left| {{V_{RM}}} \right| = \sqrt {{{\left( { - 10} \right)}^2} + {3^2}} \\
= \sqrt {109} \;{\rm{km/h}}
\end{array}\]
Therefore, the velocity of the rain with respect to man is \[\sqrt {109} \;{\rm{km/h}}\] , and the correct option is (C).
Additional information:
We know that after the resultant velocity is calculated, the direction of the velocity is also of utmost importance. The line of the resultant of two vectors is close to the vector with large magnitude, and the direction of the resultant velocities and vectors are generally calculated with the positive direction of x-axis in radians or degree. If the angle is taken in an anti-clockwise order from the positive x-axis, then the value of the angle is positive and vice versa.
Note: We must follow a proper sign convention for determining the relative velocities and taking the directions appropriately. The velocity in the positive x-direction is taken as positive, and the velocity in the positive y-direction is taken as positive. Rest in all the other options the directions are negative.
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