A person walking in a straight line, covers half of the distance to be travelled with a speed of v_(0). For half of the time, required to complete the remaining distance, he walks with a speed of v_(1)and for the remaining half time his speed is v_(2)What is the person's average speed during this complete walk ?

A person walking in a straight line, covers half of the distance to be

1.	A person walking in a straight line, covers half of the distance to be travelled with a speed of v_(0). For half of the time, required to complete the remaining distance, he walks with a speed of v_(1)and for the remaining half time his speed is v_(2)What is the person's average speed during this complete walk ?
Answer» Solution :SUPPOSE, the total walking DISTANCE is d, time taken to travel half of the distance `((d)/(2))` is `t _(1)` and time taken to travel remaining half distance is 2t. Now, distance TRAVELLED = average speed `xx` time `therefore d/2 = v _(0) t _(1)` and HENCE `t _(1) = (d)/(2v _(0)) ""...(1)` and `d/2 = v _(1) t + v _(2)t` (for remaining half distance) `= (v_(1) +v_(2))t ` `therefore 2t = (d)/(v _(1) + v _(2))` `(because t = 2t` for that half distance) ...(ii) Thus, tota time taken to walk total distance d is `t +2t.` Now the person.s average speed `lt v GT = (d)/(t _(1) + 2t)` `= (d)/(((d)/(2 v _(0))) +((d)/( v _(1) + v _(2))))` (From eq. (i) & (ii)) `therefore lt v gt = (2 v _(0) (v _(1) + v _(2)))/( v _(1) + v _(2) + 2v _(0))`