Question

A man running on a horizontal road at 8 m/s finds rain falling vertically. If he increases his speed to 12 m/s, he finds that drops make ${{30}^{\circ }}$ angle with the vertical. Find velocity of rain with respect to the road.
A: $4\sqrt{7}m/s$
B: $8\sqrt{2}m/s$
C: $7\sqrt{3}m/s$
D: $8.32m/s$

Answer 1

.This is an example of the motion that occurs in two dimensions. Hence, when we consider each element, we have to look into its horizontal and vertical components as well. Further we have to frame the connection between both the velocities by involving the concept of relative velocity. Using these methods we will be able to solve the problem.

Complete step by step answer:
From the question we are supposed to consider three main entities; velocity of the man, velocity of the rain and the velocity of the rain with respect to the man.
Velocity of rain:

& \overrightarrow{{{v}_{r}}}=a\hat{i}+b\hat{j} \\\ & \\\ \end{aligned}$$ Velocity of man: $\overrightarrow{{{v}_{m}}}$ Velocity of rain with respect to man: $\overrightarrow{{{v}_{rm}}}$ According to first condition, when the man runs with a speed of 8m/s, he sees the rain falling vertically. It is represented by $\begin{aligned} & \overrightarrow{{{v}_{m}}}=8\hat{i} \\\ & \overrightarrow{{{v}_{rm}}}=0\hat{i}+b\hat{j} \\\ & \overrightarrow{{{v}_{r}}}-\overrightarrow{{{v}_{m}}}=b\hat{j} \\\ & a\hat{i}+b\hat{j}-8\hat{i}=b\hat{j} \\\ & (a-8)\hat{i}+b\hat{j}=b\hat{j} \\\ & a-8=0 \\\ & a=8 \\\ \end{aligned}$ According to the second condition, when the velocity of the man is increased to 12m/s , the rain falls by making an angle of ${{30}^{\circ }}$ angle with the vertical. It is represented by $\begin{aligned} & \overrightarrow{{{v}_{m}}}=12\hat{i} \\\ & \overrightarrow{{{v}_{rm}}}=\overrightarrow{{{v}_{r}}}-\overrightarrow{{{v}_{m}}} \\\ & \overrightarrow{{{v}_{rm}}}=(a\hat{j}+b\hat{j})-12\hat{i} \\\ & \overrightarrow{{{v}_{rm}}}=(a-12)\hat{i}+b\hat{j} \\\ & \\\ \end{aligned}$ Since it makes ${{30}^{\circ }}$ angle with the vertical, $\begin{aligned} & \dfrac{b}{a-12}=\tan {{30}^{\circ }} \\\ & \dfrac{b}{4-12}=\dfrac{1}{\sqrt{3}} \\\ & \Rightarrow b=-\dfrac{4}{\sqrt{3}} \\\ \end{aligned}$ As we’ve got the values of a and b, the velocity of rain with respect to road is $\begin{aligned} & \overrightarrow{{{v}_{r}}}=8\hat{i}-\dfrac{4}{\sqrt{3}}\hat{j} \\\ & \therefore |\overrightarrow{{{v}_{r}}}|=\sqrt{{{(8)}^{2}}+(-{{\dfrac{4}{\sqrt{3}}}^{2}})}=8.32m/s \\\ \end{aligned}$ **Hence, option D is the correct answer among the given options.** **Note:** The sign of the relative velocity shows variations in accordance to the direction of the motion. If both the bodies are in the same directions, we get a positive value. Else it is a negative value. The students should read the question well and get an idea about the sign of the relative velocity.