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

Let the total distance be $$2d$$.

First half (distance $$d$$): Speed = 6 m/s, so time = $$\frac{d}{6}$$.

Second half (distance $$d$$): Covered in two equal time intervals. Let each interval be $$t$$.

Distance in first interval = $$9t$$, distance in second = $$15t$$.

Total: $$9t + 15t = 24t = d$$, so $$t = \frac{d}{24}$$.

Total time for second half = $$2t = \frac{d}{12}$$.

Total time = $$\frac{d}{6} + \frac{d}{12} = \frac{2d + d}{12} = \frac{3d}{12} = \frac{d}{4}$$.

Average speed = $$\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{d/4} = 8$$ m/s.

The correct answer is Option 2: 8 m/s.