Question

A man traversed half the distance with a velocity `v_(0)`. The remaining part of the distance was covered with velocity `V^(1)`. For half the time and with velocity `v_(2)` for the other half of the time . Find the average speed of the man over the whole time of motion.
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Answer 1

Time to cover the distance `AB`,
`T_(AB)=(s)/(v_(0))` …(i)
Let the time taken to cover the distance `BC` be`t_(BC)`
Given: `(t_(AB))/(2)=(s_(1))/(v_(1))` …(ii)
`S_(1)`= dismilarly, covered during first halt of time
From (ii) and (iv), `s_(1)=v_(2) (t_(AB))/(2)` ..(iii)
Similarly, `s_(2)=v_(2)(T_(AB))/(2)` ...(iv)
From (iii) and (iv),
`s_(1)+s_(2)=s=(v_(1)+v_(2))/(2) t_(BC)`
Hence, `t_(BC)=(2s)/(v_(1)+v_(2))`
Hence, average velocity `v_(AC)=(2s)/(t_(AB)+t_(BC))`
`v_(AC)=(2s)/((s)/(v_(0))+(2s)/((v_(1)+v_(2)))) rArr v_(AC)=(2v_(0)(v_(1)+v_(2)))/((v_(1)+v_(2)+2v_(0)))`.