Question 6 of 308

A particle moving in a straight line covers half the distance with speed $6 \mathrm{~m} / \mathrm{s}$. The other half is covered in two equal time intervals with speeds $9 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$ respectively. The average speed of the particle during the motion is :

A$10 \mathrm{~m} / \mathrm{s}$B$8 \mathrm{~m} / \mathrm{s}$C$9.2 \mathrm{~m} / \mathrm{s}$D$8.8 \mathrm{~m} / \mathrm{s}$

Show answer and solution▾

Answer: (B) $8 \mathrm{~m} / \mathrm{s}$

$\begin{aligned} & \mathrm{BD} \Rightarrow \mathrm{S}=9 \mathrm{t}+15 \mathrm{t}=24 \mathrm{t} \\ & \mathrm{AB} \Rightarrow \mathrm{S}=6 \mathrm{t}_1=24 \mathrm{t} \Rightarrow \mathrm{t}_1=4 \mathrm{t} \\ & \begin{aligned} \lt \text { speed }\gt & =\frac{\text { dist. }}{\text { time }}=\frac{48 \mathrm{t}}{2 \mathrm{t}+\mathrm{t}_1} \\ & =\frac{48 \mathrm{t}}{2 \mathrm{t}+4 \mathrm{t}} \Rightarrow \frac{48 \mathrm{t}}{6 \mathrm{t}} \Rightarrow 8 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$