Question

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

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

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, $v_{1}=7 \,m / s$
Body travels second half of the distance $(B$ to $C)$
in first half time with velocity, $v_{2}=14\, m / s$
and in the second half time with velocity, $v_{3}=21 m / s$
Let the time taken to travelled from $A$ to
$B=t$ second Now, distance covered from $A$ to $B=d_{A B}$
$\therefore $ Distance, $d=$ Velocity $\times$ Time
$\therefore d_{A B}=7 t$ ....(i)
Now, distance covered from $B$ to $C=d_{B C}$
$\therefore $ Average velocity of the body $=\frac{v_{2}+v_{3}}{2}$
or $v^{\prime}=\frac{14+21}{2}=\frac{35}{2} m / s$
$\therefore d_{B C}=v^{\prime} \times t^{\prime} \Rightarrow d_{B C}=\frac{35}{2} t^{\prime}$ ....(ii)
$\therefore $ Distance travelled by the body from point $A$ to $B$
$=$ distance travelled by the body from point $B$ to $C$.
$d_{A B}=d_{B C}$
From Eqs. (i) and (ii), we get
$7 t=\frac{35}{2} t^{\prime} \text { or } t^{\prime}=\frac{2}{5} t$ ....(iii)
Now, the average velocity from $A$ to $C$, for finding distance $d_{A C^{\prime}}$
$v=\frac{v_{1}+v_{2}+v_{3}}{3}=\frac{42}{3}=14 m / s$ ....(iv)
$\therefore $ Distance travelled from
$A$ to $C$, $d_{A C}=v \times t$
or $d_{A C}=14 t \ldots( v ) [$ From Eq. (iv) $]$
Total time taken from $A$ to $C$,
$T=t+t^{\prime}$
$T=t+\frac{2 t}{5}$
or $ T=t+\frac{\Delta}{2} [\therefore $ From Eq. (iii) $]$
$T=\frac{7 t}{5}$ .....(iv)
Now, average velocity during the whole journey $($ From $A$ to $C)$,
$v_{\text {avg }}=\frac{d_{A C}}{T} \text { or } v_{\text {avg }}=\frac{14 \,t}{7\, t} \times 5$
[ $\therefore $ From Eqs. (v) and (vi) ]
or $v_{\text {avg }}=10 \,m / s$