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

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

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Given, a body travelling along a straight line path. Body travels first half of the distance $(A$ to $B$) with velocity, $v_1=7 \mathrm{~m} / \mathrm{s}$ Body travels second half of the distance ( $B$ to $C$) in first half time with velocity, $v_2=14 \mathrm{~m} / \mathrm{s}$ and in the second half time with velocity, $v_3=21 \mathrm{~m} / \mathrm{s}$ 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 $\therefore \quad d_{A B}=7 t$...(i) Now, distance covered from $B$ to $C=d_{B C}$ $\therefore$ Average velocity of the body $=\frac{v_2+v_3}{2}$ or $\therefore \quad d_{B C}=v^{\prime} \times t^{\prime} \Rightarrow d_{B C}=\frac{35}{2} t^{\prime}$...(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 $d_{A C}$, $v=\frac{v_1+v_2+v_3}{3}=\frac{42}{3}=14 \mathrm{~m} / \mathrm{s}$...(iv) $\therefore$ Distance travelled from $A$ to $C$, $\begin{aligned} & d_{A C}=v \times t \\ & \text{or } d_{A C}=14 t \quad \text{...(v)[From Eq. (iv)]} \end{aligned}$ Total time taken from $A$ to $C$, $\begin{aligned} & T=t+t^{\prime} \\ & \text{or } T=t+\frac{2 t}{5} \quad [\therefore \text{ From Eq. (iii)]}\\ & T=\frac{7 t}{5} \quad \ldots (vi) \end{aligned}$ 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 $v_{\text {avg }}=10 \mathrm{~m} / \mathrm{s}$

A body travelling along a straight line path travels first half of the distance with a velocity \(7…

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