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Average Velocity in 2D Motion

A particle mo...

Question

A particle moves half the distance with velocity $u$ . Rest of the half distance is covered with velocities $v_{1}$ and $v_{2}$ in equal intervals of time. Find the average velocity of the particle.

\frac{2 u (v_{1} + v_{2})}{2 u + v_{1} + v_{2}}

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\frac{2 u (v_{1} + v_{2})}{u + v_{1} + v_{2}}

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\frac{2 u (v_{1} + v_{2})}{u + 2 (v_{1} + v_{2})}

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\frac{u + \frac{v_{1} + v_{2}}{2}}{2}

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Solution

The correct option is A $\frac{2 u (v_{1} + v_{2})}{2 u + v_{1} + v_{2}}$
Motion of the particle is shown above.
Let the total distance covered by the particle be $s$ .
For second half motion :
Distance covered $x + y = \frac{s}{2}$ .....(1)
Equal interval of time is taken to cover $x$ and $y$ distances with speed $v_{1}$ and $v_{2}$ , respectively.
$\Rightarrow \frac{x}{v_{1}} = \frac{y}{v_{2}}$
$\Rightarrow x v_{2} = y v_{1}$
$\Rightarrow (\frac{s}{2} - y) v_{2} = y v_{1}$ (using 1)
$\Rightarrow \frac{s}{2} v_{2} = y (v_{1} + v_{2})$
$\Rightarrow y = \frac{s v_{2}}{2 (v_{1} + v_{2})}$
From (1), we get $x + \frac{s v_{2}}{2 (v_{1} + v_{2})} = \frac{s}{2}$
$\Rightarrow x = \frac{s v_{1}}{2 (v_{1} + v_{2})}$

Time taken to cover second half motion, $t = \frac{x}{v_{1}} + \frac{y}{v_{2}} = \frac{s}{(v_{1} + v_{2})}$
Time taken to cover first half motion, $t^{'} = \frac{s / 2}{u} = \frac{s}{2 u}$
Now average velocity $V_{a v g} = \frac{T o t a l d i s t a n c e}{T o t a l t i m e}$
$∴$ $V_{a v g} = \frac{s}{\frac{s}{2 u} + \frac{s}{(v_{1} + v_{2})}}$
$⟹$ $V_{a v g} = \frac{2 u (v_{1} + v_{2})}{v_{1} + v_{2} + 2 u}$

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