Average velocity is total displacement divided by total time, not the simple average of the two speeds. Because the path is straight and the motion is uniformly accelerated, the displacement equals the average of the initial and final velocities multiplied by time: \(s = \frac{u + v}{2} \times t\).

Here \(u = 10\) m/s and \(v = -2\) m/s (the negative sign shows the object reversed direction). So \(s = \frac{10 + (-2)}{2} \times 5 = \frac{8}{2} \times 5 = 4 \times 5 = 20\) m.

Average velocity \(= \frac{\text{displacement}}{\text{time}} = \frac{20}{5} = 4\) m/s.

Hence, the magnitude of the average velocity is 4 m/s. Note that 6 m/s would be the plain average of the speeds 10 and 2, which is wrong because velocity is a vector and the directions are opposite; 8 m/s and 5 m/s do not follow from the displacement calculation.