I couldn't solve this.-
Rain is falling vertically with a speed of 30m/s.A woman on a bicycle is travelling with a speed of 10m/s in the north to south direction.In what direction should she hold her umbrella in order to protect her from rain?
With what speed does the rain strike the umbrella?
Plz make me understand by drawing diagram.
1here the umbrella may be held in the direction in which the rain appears to women
so vrain w.r.t. women = vrain w.r.t ground -- vwomen wrt ground
= -30k -10i
the direction in which this thing is will give you the direction and magnitude is speed with which ti hits
1I didn't understand this sum.I will b grateful if u plz make me understand this sum.
39The first thing to do in such problems is to reserve the vertical axes for the direction of rainfall and the horizontal axes for observer movement.
Hence we are given that the actual direction of rainfall is vertically downwards with a speed 30m/s, -30j when written in vector form.
Let the left direction be north and the right direction be south. Hence the woman is traveling in the positive direction of the X-axis, which is the positive "i" direction. Her speed is 10i m/s in vector form.
First we need the direction of rainfall relative to observer movement.
Vrw = Vr + (-Vw) = -30j - 10i m/s
We need to find the angle A.
Now to find magnitude of Vrw,
Vrw = √V²r + V²w + 0 as cos90° = 0.
=> Vrw = √100 + 900 = √1000
From here, you can use some easy trigonometry to find angle A. When you find it, the answer to direction will be - The woman should hold her umbrella A° south of vertical as that forward direction is south.
The speed by which rain strikes umbrella is √1000 m/s as it is the magnitude of Vrw(not sure about this one, check).
