Rain is falling vertically with a speed of ` 30 ms^(-1)` . A woman rides a bicycle with a speed of ` 10 ms^(-1)` in the North to South direction. What is the direction in which she should hold her umbrella ?

Explanation

In Fig. 2 (NCT). 20, the rain is falling along ` OA` with speed ` 30 ms^(-1)` and woman rider is moving along ` OS` with speed ` 10 ms ^(-1)`i. e. ` OA = 30 ms^(-1) & OB = 10 ms^(-1)` . The woman rider can protect herself from the rain if she holds her umberella in the direction of relative velocity of rain w.r.t. woman. To do so apply equal and opposite velocity of woman on the rain i.e. impress thevelocity ` 10 ms ^(-1)` due
North on rain which is represented by ` vec (OC)` . Now the relative velocity or rain ` w.r.t. woman will be represented by diagonal ` vec (OD)` of parallelogram ` OADC`, If ` /_AOD = theta`, then in ` Delta OAD`,
` tan = (AD)/(OA) = (OC)/(OA) = (10)/(30) =0.333 = tan 18^@ 26'`
or ` beta = 18^@ 26 ' ` with vertical in forwward direction.
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