The rain is falling vertically with a speed of 12 m/s, and the woman is moving with a speed of 12 m/s in the east to west direction.
In this case, the rain appears to come at an angle relative to the woman.
To calculate the direction in which she should hold her umbrella, we can use the concept of relative velocity.
The velocity of the rain relative to the woman is the vector sum of the downward velocity of the rain and the horizontal velocity of the woman.
Both velocities are equal in magnitude (12 m/s), and they are perpendicular to each other.
The resultant velocity of the rain relative to the woman will make an angle with the vertical direction.
The angle \(\theta\) can be calculated using the relation: \[ \tan \theta = \frac{v_{\text{horizontal}}}{v_{\text{vertical}}} \] Since both velocities are equal, we have: \[ \tan \theta = \frac{12}{12} = 1 \] \[ \theta = \tan^{-1}(1) = 45° \]
Thus, the woman should hold her umbrella at an angle of 45° towards the west (in the opposite direction of her motion).
The correct answer is (D) : 45° towards West.
To find the direction in which the woman should hold her umbrella, we need to consider the velocity of the rain relative to the woman. Since the rain is falling vertically with a speed of 12 m/s and the woman is moving with the same speed of 12 m/s in the east to west direction, we can treat this as a vector addition problem.
The velocity of the rain can be decomposed into two components:
- The vertical component (\( V_{\text{rain, vertical}} \)) is 12 m/s downward.
- The horizontal component (\( V_{\text{rain, horizontal}} \)) is the velocity of the woman, which is 12 m/s in the westward direction.
The resultant velocity of the rain relative to the woman is the vector sum of these two components. Using the Pythagorean theorem:
\[V_{\text{resultant}} = \sqrt{(12^2 + 12^2)} = \sqrt{288} = 12\sqrt{2} \, \text{m/s}\]The angle \( \theta \) at which the umbrella should be tilted is given by:
Thus, the woman should hold her umbrella at \( 45^\circ \) towards the west.