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Question
Rain is falling vertically with a speed of 30 m s–1. A woman rides a bicycle with a speed of 10 m s–1in the north to south direction. What is the direction in which she should hold her umbrella?
Solution 1
The described situation is shown in the given figure.
Here,
vc = Velocity of the cyclist
vr = Velocity of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.
`vecv = vec(v_r) + (-vecv_c)`
Resultant, |v| = `sqrt(v_(r^2)+ v_(c^2))`
|v| = `sqrt(30^2+10^2)`
|v| = `sqrt(900+100)`
|v| = `sqrt1000 = 10sqrt10m`
`tan theta = v_c/v_r = 10/30`
`theta = tan^(-1)(1/3)`
`theta = tan^(-1)(0.333) ~~18^@`
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical
Solution 2
The situation has been demonstrated in the figure below. Here `vecv = 30 "ms"^(-1)` is the rain velocity in vertically downward direction and `vecv_c = 10 ms^(-1)` is the velocity of cyclist women in horizontal plane from north N to south S.
∴Relative velocity of rain w.r,t cyclist `vecv_(rc)` subtends an angle β with verticle such that
`tan beta = |vecv_c|/|vecv_r| = 10/30 = 1/3`
`:. beta = tan^(-1)(1/3) - 18^@26`
Hence the woman should hold her umberlla at `18^@26` south of verticle
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