Question

Rain is falling vertically downwards with a speed of $4km{h^{ - 1}}$. A girl moves on a straight road with a velocity of $3km{h^{ - 1}}$. The apparent velocity of rain with respect to the girl is
(A) $3km{h^{ - 1}}$
(B) $4km{h^{ - 1}}$
(C) $5km{h^{ - 1}}$
(D) $7km{h^{ - 1}}$

Answer 1

Hint
The apparent or the relative velocity of the rain with respect to the girl is given by the vector sum of the velocity vector of the rain and the negative of the velocity vector of the girl, where the angle between the 2 vectors will be $90^\circ $.
In this solution we will be using the following formula,
$\Rightarrow {\vec V_{AB}} = {\vec V_A} - {\vec V_B}$
Where ${\vec V_A}$ and ${\vec V_B}$ are 2 velocity vectors and ${\vec V_{AB}}$ is the velocity of A with respect to the velocity of B.

Complete step by step answer
In the problem we are given the velocity of rain as, ${\vec V_R} = 4km{h^{ - 1}}$ in the downward direction. And the velocity of the girl on the straight road is given by, ${\vec V_G} = 3km{h^{ - 1}}$.
So the apparent velocity of the rain with respect to the girl can be denoted by ${\vec V_{RG}}$. This will have a value given by,
$\Rightarrow {\vec V_{RG}} = {\vec V_R} - {\vec V_G}$
Now we can also write this as,
$\Rightarrow {\vec V_{RG}} = {\vec V_R} + \left( { - {{\vec V}_G}} \right)$
So the apparent velocity is the vector sum of the velocity of rain and the negative of the velocity of the girl given by, $ - {\vec V_G} = - 3km{h^{ - 1}}$
According to the parallelogram law for the sum of vectors, we can write,
$\Rightarrow {\vec V_{RG}} = \sqrt {{{\left| {{{\vec V}_R}} \right|}^2} + {{\left| { - {{\vec V}_G}} \right|}^2} + 2{{\vec V}_R} \cdot \left( { - {{\vec V}_G}} \right)\cos \theta } $
Now the angle between the rain and the girl is $90^\circ $. So the cosine of the angle is , $\Rightarrow \cos 90^\circ = 0$. So substituting all the values we get,
$\Rightarrow {\vec V_{RG}} = \sqrt {{{\left| 4 \right|}^2} + {{\left| { - 3} \right|}^2} + 0} $
Hence, on doing the squares,
$\Rightarrow {\vec V_{RG}} = \sqrt {16 + 9} $
On adding we get the value as,
$\Rightarrow {\vec V_{RG}} = \sqrt {25} = 5km{h^{ - 1}}$
So the apparent velocity of the rain as seen by the girl will be ${\vec V_{RG}} = 5km{h^{ - 1}}$
So the correct answer is option (C).

Note
The relative velocity of one body with respect to another is the velocity with which an observer sees a body move being in the rest frame of the other body. In case of one dimensional motion, the relative velocity is found by the sum or the difference of the motion of the two bodies according to their direction.