A particle moving in a straight line covers half the distance with a speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speeds of 4.5 m/s and 7.5 m/s respectively. The average speed (in m/s) of the particle during this motion is 

Let the total distance be S. The time taken to travel the first half, i.e., $\frac{\mathrm{S}}{2}$is 
${\mathrm{t}}_1=\frac{\mathrm{S}/2}{3}=\frac{\mathrm{S}}{6}$
Let t₂ be the time taken to cover a distance S₁ with speed 4.5 m/s and t₃ that to cover distance S₂ with speed 7.5 m/s.
Then ${\mathrm{S}}_1=4.5{\mathrm{t}}_2\text{ and }{\mathrm{S}}_2=7.5 {\mathrm{t}}_3$
Now, ${\mathrm{S}}_1+{\mathrm{S}}_2=\frac{\mathrm{S}}{2} \mathrm{and} {\mathrm{t}}_2={\mathrm{t}}_3 (\mathrm{given})$
Therefore $\frac{\mathrm{S}}{2}={\mathrm{S}}_1+{\mathrm{S}}_2$
        $=(4.5+7.5){\mathrm{t}}_2\text{ or }{\mathrm{t}}_2=\frac{\mathrm{S}}{12}.$
$\therefore$Total time taken $\mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2=\frac{\mathrm{S}}{6}+\frac{\mathrm{S}}{12}=\frac{\mathrm{S}}{4}$
$\therefore$ Average speed $=\frac{\text{ total distance }}{\text{ total time }}=\frac{\mathrm{S}}{\mathrm{S}/4}$
                              $=4 \mathrm{m}/\mathrm{s}$