50

Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. 
A beats a drum and B hears the sound t₁ time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum t₂ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.


 The beats are a unit of time.
t = d/v

In the first experiment, the wind is assisting the speed of sound. The total speed is v + u. Therefore the time = x / (v + u)
11 = x / (v + u)
11v + 11u = x

In the second experiment, the wind is going against the travel of direction. v - u
The equation you get is
12v - 12u = x

The two x's are an equal distance So you can begin by equating the two distances.
12v - 12u = 11v + 11u
v = 23 u

Now you can go to one of the original equations.
12v - 12u = x
12*23u - 12u = x
12u (23 - 1) = x
12u * 22 = x
u = x / 264

v = 23*u
v = 23 * x/264

Answer: In the first case resultant velocity = u+v  

So the time taken t₁ = x/(u+v)  

→ u+v = x/t₁  ..................(A)   

In the second case resultant velocity = v-u   

Now the time taken = t₂ = x/(v-u)   

→ v-u = x/t₂    ................(B)  

Adding the equations (A)  and  (B)  

2 v = x/t₁ + x/t₂ 

→ v = x/2  * (1/t₁+1/t₂)   It is the velocity of sound in still air.

 Subtracting (B) and (A) we get 

2u = x/t₁ - x/t₂   

→ u = x/2  * (1/t₁ - 1/t₂)      It is the velocity of wind . 

the velocity of sound and wind are added vectorially.
therefore, in the first case,

v + u = x/t1 ........................ (1)

v - u = x/t2 ...........................(2)
on adding, we get
v = x (1/t1 + 1/t2) /  2
on subtracting,
u = x (1/t1 - 1/t2) /  2

in the second case,

the resultant velocity and u are at 900 to each other
therefore,if u make a rt. triangle, 
                                   
v acts as hypotenuse and u as a leg and resultant vel. towards B makes the other leg. therefore,
Resultant vel. =   v2 - u2 
time taken =     x      
                            v2 - u2 

49

Q 49. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

I will form the diagram for the given problem.

Let us first assign the unknown variables.
t = Total time of flight (unknown)
A = Angle of destination vector from north = 30∘
B = Angle of airplane's propulsion, measured from East (to find)
V = Final Velocity (Unknown)
u = Air Velocity = 20ms−1
s = Airplane Velocity = 150ms−1

The air velocity only provides a northward displacement, ie the Eastward displacement should come from the East component of the plane's velocity.

The eastward displacement is 500×cos(90∘−A)km
The East component of the airplane's velocity is s×cos(B)

So the equation is 
500×cos(90∘−A)=s×cos(B)×t

You will obtain an equation in B and t.

Now, the northward displacement is due to both the airplane's propulsion and the wind velocity.
The net northward component of velocity is
s×sin(B)+u

The distance is 
500×sin(90∘−A)

Thus the equation is
500×sin(90∘−A)=(s×sin(B)+u)×t

You have another equation in B and t.

Solve for B