Question

An object covers first half of the total distance with speed V1 and second half of the total distance with speed V2. If average speed for the entire trip is V3, then prove that 1/v3= 1/2 [1/v1 + 1/v2].​

Answer 1

Answer:

Just take the reciprocal.

Explanation:

V average=v1+v2/2

that is v3=v1+v2/2 -(1)

Take the reciprocal of (1)

that is 1/v3=1/2[1/v1+1/v2]

Answer 2

Answer:

Hence, it is proved that  $V_{3} = \frac{1}{2}(\frac{1}{V_{1} }+\frac{1}{V_{2} })$

Explanation:

Given:

First half of the total distance with speed = V₁

Second half of the total distance with speed = V₂

Average speed = V₃

Formula

$Velocity = \frac{displacement}{Time}$  ...................................(1)

$Time = \frac{Displacement}{Velocity}$ .......................................(2)

Solution:

The half distance (x) covered with the speed V₁ in t₁ time.

$t_{1} = \frac{x}{V_{1} }$ .....................................................(3)

Another half distance (x) covered with speed V₂ in time t₂

$t_{2} = \frac{x}{V_{2} }$    ................................................(4)

$Average velocity = \frac{Total distance}{Total time}$......................(5)

$Total time = t_{1} + t_{2}$

                 $= \frac{x}{V_{1} }+ \frac{x}{V_{2} }$

Take LCM for equation (6) we get

$Total time = \frac{V_{2} x + V_{1} x}{V_{1}V_{2} }$     ...................................(6)

Total distance = x + x

                     = 2 x ......................................(7)

Substitute equation (6) and (7) in equation (5) we get

$Average velocity V_{3} = \frac{2x}{\frac{V_{2} x + V_{1} x}{V_{1} V_{2} } }$ .......................(8)

Take Reciprocal for equation (8)

$\frac{1}{V_{3} } = \frac{1}{2} [\frac{1}{V_{1} }+\frac{1}{V_{2} } ]$

Hence it is proved that $\frac{1}{V_{3} } = \frac{1}{2}[ \frac{1}{V_{1} }+ \frac{1}{V_{2} }]$

#SPJ2