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Question: A man is cycling at \(4\,m/s\) . To him rain appears at \({30^ \circ }\) to vertical. If he doubles ...

A man is cycling at 4 m/s4\,m/s . To him rain appears at 30∘{30^ \circ } to vertical. If he doubles his velocity, rain appears to fall at 60∘{60^ \circ } to vertical. Find the velocity of the rain.

Explanation

Solution

Hint-
Use vector addition to find the resultant velocity or the velocity of rain with respect to man. Then take tan of the given angles in both cases .

On comparing the equation that we get in case one and case two we can find the x component and y component of the velocity of rain. Then find the magnitude of the vector of velocity of rain to get the final answer.

Step by step solution:
It is given that the velocity of the man is 4 m/s4\,m/s. Let us denote the velocity of the man as vM{v_M} .
Thus,
vM=4 m/s{v_M} = 4\,m/s
At this velocity the rain appears to fall at an angle 30∘{30^ \circ } at vertical.
When he increases his speed to double the initial value rain appears to fall at an angle 60∘{60^ \circ } at vertical.
Let this velocity be denoted as vM′{v_M}^\prime since it is double the initial velocity we can write
vM′=2×4 m/s=4 m/s{v_M}^\prime = 2 \times 4\,m/s = 4\,m/s
Now let us denote the velocity of rain as vR{v_R} .
In a general form we can write this as
vR=a i^+bj^{v_R} = a\,\widehat i + b\widehat j
Here a denotes the x component and b denotes the y component.
Let us consider the first case.

The velocity of rain with respect to man is denoted as vRM{v_{RM}} .
vM=4 i^{v_M} = 4\,\widehat i
Using the law of vector addition we can write
vRM=vR−vM{v_{RM}} = {v_R} - {v_M}
That is,
vRM=a i^+bj^−4 i^=(a−4)i^+bj^{v_{RM}} = a\,\widehat i + b\widehat j - 4\,\widehat i = \left( {a - 4} \right)\widehat i + b\widehat j
The angle made with the vertical is given as 30∘{30^ \circ }
Thus tan 30∘{30^ \circ } will be
tan⁡30∘=ba−4\tan {30^ \circ } = \dfrac{b}{{a - 4}}.................(1)
Similarly in the second case when the velocity doubles we have the angle made with the vertical as 60∘{60^ \circ } .
Here, velocity of man vM′=4 m/s{v_M}^\prime = 4\,m/s
Let the velocity of rain with respect to man be denoted as vRM′{v_{RM}}^\prime .
Then ,
vRM′=vR−vM′{v_{RM}}^\prime = {v_R} - {v_M}^\prime
That is,
vRM′=a i^+bj^−8i^=(a−8)i^+bj^{v_{RM}}^\prime = a\,\widehat i + b\widehat j - 8\widehat i = \left( {a - 8} \right)\widehat i + b\widehat j
Since the angle made with horizontal is 60∘{60^ \circ } .
We can write ,
tan⁡60∘=ba−8\tan {60^ \circ } = \dfrac{b}{{a - 8}} ……………...(2)
Now let us divide equation 1 by equation 2.
tan⁡30∘tan⁡60∘=ba−4ba−8\dfrac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} = \dfrac{{\dfrac{b}{{a - 4}}}}{{\dfrac{b}{{a - 8}}}}
⇒133=a−8a−4\Rightarrow \dfrac{{\dfrac{1}{{\sqrt 3 }}}}{{\sqrt 3 }} = \dfrac{{a - 8}}{{a - 4}}
⇒13=a−8a−4\Rightarrow \dfrac{1}{3} = \dfrac{{a - 8}}{{a - 4}}
⇒a−4=3a−24\Rightarrow a - 4 = 3a - 24
⇒2a=20\Rightarrow 2a = 20
∴a=10\therefore a = 10
Now let us substitute this value in equation 1.
3=b10−8\sqrt 3 = \dfrac{b}{{10 - 8}}
⇒3×2=b\Rightarrow \sqrt 3 \times 2 = b
∴b=3.46\therefore b = 3.46
Thus we get ,
vR=a i^+bj^=10i^+3⋅46j^{v_R} = a\,\widehat i + b\widehat j = 10\widehat i + 3 \cdot 46\widehat j
The magnitude of this velocity can be found by as
∣vR∣=102+3⋅462\left| {{v_R}} \right| = \sqrt {{{10}^2} + 3 \cdot {{46}^2}}
∴∣vR∣=10⋅58 m/s\therefore \left| {{v_R}} \right| = 10 \cdot 58\,m/s
This is the value of velocity of rain.

Note: While doing vector addition take care of the direction of vectors .We will get the resultant vector by adding the velocity of rain vR{v_R} with the negative of the velocity of man .because the resultant vector is obtained when the other sides are joined by head to tail and resultant vector is drawn from the end of free tail to the end of free head. For this we need to take the negative direction of velocity of man.