Question

A body traversed half of the distance with a velocity ${v_0}$. The remaining part ${v_1}$ for half of the time and with the velocity ${v_2}$ for the other half of the time. Find the average velocity of the body over the whole journey.
(A) $\dfrac{{2{v_0}({v_1} + {v_2})}}{{(2{v_0} + {v_1} + {v_2})}} \\ $
(B) $\dfrac{{3{v_0}({v_1} + {v_2})}}{{(2{v_0} + {v_1} + {v_2})}} \\ $
(C) $\dfrac{{2{v_0}({v_1} + {v_2})}}{{(3{v_0} + {v_1} + {v_2})}} \\ $
(D) $\dfrac{{{v_0}({v_1} + {v_2})}}{{(2{v_0} + {v_1} + {v_2})}} $

Answer 1

Hint :
In order to solve this problem, first we have to calculate the time taken by body to travel first half distance with ${v_0}$ by using following expression
Time $ = $ distance $/$ velocity
Now, we will calculate the time taken in the next half distance. For that we use following expression
Distance $ = $ speed $ \times $ time
Because for the next half distance, time is divided.
At last putting the values in average velocity. Expression, we get desire solution i.e.,
Average velocity $ = $ Total distance travelled by body $/$ Total time taken by body

Complete step by step solution :
Let the total distance travelled by the body is d.
Given that half of the distance travelled with velocity ${v_0}$. So, time taken to travel half of the distance is given by following expression
Time $ = $ distance $/$ velocity
${t_1} = \dfrac{d}{{2{v_0}}}$ …..(1)
Now, let remaining half distance is travelled in time ${t_2}$ and given that $\dfrac{{{t_2}}}{2}$ be the time for which body travels with speed ${v_1}$ and next $\dfrac{{{t_2}}}{2}$ be the time for which body travels with ${v_2}$.
So, using distance-time formula
Distance $ = $ speed $ \times $ time
$\dfrac{d}{2} = {v_1} \times \dfrac{{{t_2}}}{2} + \dfrac{{{v_2}{t_2}}}{2}$

$d = {v_1}{t_2} + {v_2}{t_2}$
${t_2}({v_1} + {v_2}) = d$
${t_2} = \dfrac{d}{{{v_1} + {v_2}}}$ …..(2)
We know that the average speed is given as
Average speed $ = $ total distance $/$ total time
${v_{avg}} = \dfrac{d}{{{t_1} + {t_2}}}$
From equation 1 and 2

${v_{avg}} = \dfrac{d}{{\left( {\dfrac{d}{{2{v_0}}}} \right) + \left( {\dfrac{d}{{{v_1} + {v_2}}}} \right)}}$

$ = \dfrac{d}{d}\left[ {\dfrac{1}{{\dfrac{1}{{2{v_0}}} + \dfrac{1}{{({v_1} + {v_2})}}}}} \right]$

${v_{avg}} = \dfrac{{2{v_0}({v_1} + {v_2})}}{{(2{v_0} + {v_1} + {v_2})}}$

Hence option A is the correct answer.

Note :
Many times, students may get confused between average velocity and instantaneous velocity.
Average velocity $ = $ Total displacement travelled by body $/$ Total time taken by body
${\overrightarrow v _{avg}} = \dfrac{{\Delta \overrightarrow r }}{{\Delta t}}$
When $\Delta t \to 0$ then average velocity is converted into instantaneous velocity.
So, ${\overrightarrow v _{inst}} = \dfrac{{d\overrightarrow r }}{{dt}}$
Rate of change in displacement with respect to time.