When I started preparation for jee,  I thought  Irodov  is  very difficult.  A close reading of the book reveals that it is not so provided, i repeat, one reads HCV .

Let us take 1 example.

Q. 1.1
A motorboat going downstream overcame a raft at a point A ;  t= 60 mins later it turned back and after some time passed the raft at a distance L= 6.0 km from the point A. Find the flow velocity assuming the duty of the engine to be constant.


solution :

How to solve the problem ??

Basically this requires the concept of relative motion---more specifically, relative motion in one dimension.

Let us assume that the flow of water is from left to right ( +ve dirn of x axis ).
This is downstream. The motorboat is going downstream.

At t = 0 , ( beginning ) , the raft and motorboat are at point A.  The velocity of the raft ( no engine ) is = velocity of the river.

At  t = 60 mins, motorboat ( with engine ) goes far away from raft , say , at point P ( in the +ve dirn of x axis ) , ahead of raft which proceeds only up to point B ( in the +ve dirn of x axis ) .

Clearly, time taken by the boat to reach P from A = time taken by the raft to reach B from A.

Then  the boat turns back from the point P , but raft proceeds from B to C.

Then the 2 meet at point C. The boat going upstream and raft going downstream.

Let the 2 meet at time  = t + t ₁  at point C.

So, time taken by raft to reach from B to C = time taken by boat to reach  C  from P in the reverse motion ( upstream motion )

For our visualization, we can consider a line along X axis with origin at A , then B. then C and then point P.

Now  V _a = actual velocity of boat ( with engine )

        V _b = actual velocity of raft = velocity of river.

so, during downstream ( going along +ve X axis ) ,

if V _c = relative velocity of motor wrt stream, then

                     V _c = V _a  -- V _b

                      V _a  =  V _b +   V _c

                  so time  t = AP /  V _a  =  AP / ( V _b +   V _c )

But AB = distance travelled by raft in time t = t ( V _b )

now let us focuss abt upstream when the motorboat turns back. ( going from right to left or towards -ve X axis ).

                    During upstream  V _c = V _a  +   V _b

( the boat turns back but raft goes forward )

                                so,  V _a  = V _c --  V _b 

So, PC = distance travelled by motorboat in upstream in time   t _c

              =  ( V _c --  V _b  )   t _c

BC = distance travelled by raft   in time   t _c  = ( V _b  ) t _c

now
                 AP - PC = AC = L  [ note again A, B then C and then P ]


( V _b +   V _c ) t  -- ( V _c --  V _b  )   t _c  = L


V _c t +  V _b t -- V _c  t _c  +  V _b   t _c   = L ..................(1)

again

                  AB + BC = L

or,  V _b t +  V _b   t _c = L

          V _b  = L /  ( t + t _c )   ..........................................(2)

manipulating 2 and 1 , we get
                                                V _b = L / 2t



Q.1.2

A point traversed half the distance with a velocity V ₀ . The remaining part of the velocity was covered with velocity V ₁  for half the time and with velocity V ₂ for other half of the time. Find the mean velocity of the point averaged over the whole time of motion

soln :

Basically this requires the concept of average velocity in  one dimension.

V _av = total displacement / total time

Let us again visualize points A, B and C serially along X axis from left to right.

Let the particle starts from A at t = 0 . It goes to B at t = t ₁  and then to C
 at t = t ₂ .

now,  AB = BC = L   ( say )

t₁   =  L / V ₀  ......................................( 1)

The time taken by the particle to reach C from B is t ₀ = t ₂  - t₁  

            BC = L =  V ₁  ( t ₂ - t ₁  ) / 2  +      V ₂   ( t ₂ - t ₁  ) / 2 ............(2)

V _av = total displacement / total time  = AC / t ₂ = 2L / t₂ 


from 1 and 2,    L =  V ₁  ( t ₂ - t ₁  ) / 2  +      V ₂   ( t ₂ - t ₁  ) / 2

V ₀  t ₁ = V ₁  ( t ₂ - t ₁  ) / 2  +      V ₂   ( t ₂ - t ₁  ) / 2

doing manipulation,

                V _av = 2 V ₀  (  V ₁ +  V ₂ )  /  2 V ₀  +   V ₁ +  V ₂

This result can be memorized  for short -cut.