KCET · Physics · Motion In Two Dimensions
Rain is falling vertically with a speed of \(12 \mathrm{~ms}^{-1}\). A woman rides a bicycles with a speed of \(12 \mathrm{~ms}^{-1}\) in east to west direction. What is the direction in which she should hold her umbrella?
- A \(30^{\circ}\), towards east
- B \(45^{\circ}\), towards east
- C \(30^{\circ}\), towards west
- D \(45^{\circ}\), towards west
Answer & Solution
Correct Answer
(B) \(45^{\circ}\), towards east
Step-by-step Solution
Detailed explanation
The given situation is shown below
Here, velocity of rain, \(v_{r}=12 \mathrm{~ms}^{-1}\)
and velocity of woman, \(v_{w}=12 \mathrm{~ms}^{-1}\)
The relative velocity of rain w.r.t. woman is
\(v_{r w} =\sqrt{v_{w}^{2}+v_{r}^{2}} \)
\( =\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}\)
The direction in which the woman should hold her umbrella is
\(\sin \theta =\frac{v_{w}}{v_{r w}}=\frac{12}{12 \sqrt{2}} \\ \Rightarrow \theta =\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}, \text { towards east}\)
Here, velocity of rain, \(v_{r}=12 \mathrm{~ms}^{-1}\)
and velocity of woman, \(v_{w}=12 \mathrm{~ms}^{-1}\)
The relative velocity of rain w.r.t. woman is
\(v_{r w} =\sqrt{v_{w}^{2}+v_{r}^{2}} \)
\( =\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}\)
The direction in which the woman should hold her umbrella is
\(\sin \theta =\frac{v_{w}}{v_{r w}}=\frac{12}{12 \sqrt{2}} \\ \Rightarrow \theta =\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}, \text { towards east}\)
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