Question

A body travelling along a straight line path travels first half of the distance with a velocity $7m{s}^{-1}$. During the travel time of the second half of the distance, first half time is travelled with a velocity $14m{s}^{-1}$ and the second half time is travelled with a velocity $21m{s}^{-1}.$ Then the average velocity of the body during the journey is

• A. $14m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $9m{s}^{-1}$
• D. $12m{s}^{-1}$

Answer 1

Answer: (B)

According to the question,

Given, a body travelling along a straight line path. Body travels first half of the distance $(A$ to $B)$ with velocity, $v_1=7m/s$
Body travels second half of the distance $(B$ to $C)$
in first half time with velocity, $v_2=14m/s$
and in the second half time with velocity, $v_3=21m/s$
Let the time taken to travelled from $A$ to
$B=t$ second Now, distance covered from $A$ to $B=d_{AB}$
$\therefore$ Distance, $d=$ Velocity $\times$ Time
$\therefore d_{AB}=7t$ ....(i)
Now, distance covered from $B$ to $C=d_{BC}$
$\therefore$ Average velocity of the body $=\frac{v_2+v_3}{2}$
or $v^{\prime }=\frac{14+21}{2}=\frac{35}{2}m/s$
$\therefore d_{BC}=v^{\prime }\times t^{\prime }\Rightarrow d_{BC}=\frac{35}{2}{t}^{\prime }$ ....(ii)
$\therefore$ Distance travelled by the body from point $A$ to $B$
$=$ distance travelled by the body from point $B$ to $C$.
$d_{AB}=d_{BC}$
From Eqs. (i) and (ii), we get
$7t=\frac{35}{2}{t}^{\prime }\text{ or }{t}^{\prime }=\frac{2}{5}t$ ....(iii)
Now, the average velocity from $A$ to $C$, for finding distance $d_{A{C}^{\prime }}$
$v=\frac{v_1+v_2+v_3}{3}=\frac{42}{3}=14m/s$ ....(iv)
$\therefore$ Distance travelled from
$A$ to $C$, $d_{AC}=v\times t$
or $d_{AC}=14t\dots (v)[$ From Eq. (iv) $]$
Total time taken from $A$ to $C$,
$T=t+t^{\prime }$
$T=t+\frac{2t}{5}$
or $T=t+\frac{\Delta }{2}[\therefore$ From Eq. (iii) $]$
$T=\frac{7t}{5}$ .....(iv)
Now, average velocity during the whole journey $($ From $A$ to $C)$,
$v_{\text{avg }}=\frac{d_{AC}}{T}\text{ or }{v}_{\text{avg }}=\frac{14t}{7t}\times 5$
[ $\therefore$ From Eqs. (v) and (vi) ]
or $v_{\text{avg }}=10m/s$