Question

A body travels from A to B with a velocity V1 and return back from B to A with a velocity V2. The
average speed of the body during the whole journey is


explain with steps

Answer 1

Now, let the distance between A & B be 'd'.
We're gonna use the formula, (Distance)=(Speed)(Time).
Time taken to go from A to B is t1=d/V1.
Time taken to come from B to A is t2=d/V2.
Now, you know that average velocity is (total distance)/(total time).
Then, V(average)=(2d)/(t1+t2)=$\frac{2d}{\frac{d}{V1}+\frac{d}{V2}}$. You get V(average)=2(V1)(V2)/(V1+V2)

Answer 2

Time taken to travel from A to B = AB/V1
Time for traveling from B to A = AB / V2

Total Time taken to travel AB and BA,
        = AB [ 1/V1 + 1/V2 ] = AB * (V1+V2)/V1* V2

Average speed  =  total distance / total time = 2 * AB / [ AB *  (V1+V2) / V1 * V2 ]
                 = 2 V1 * V2 / (V1 + V2 ]

Average speed will be the reciprocal of the average of the reciprocals of speeds.