A particle moves along a straight line with position s(t)=α t²-β t+γ, where α=1 ms⁻², β=6 ms⁻¹, γ=5 m. The average speed of the particle (in ms⁻¹) from t=0 to t=6 s is:

Solution & Explanation

Caution: 'average speed' uses total DISTANCE travelled (path length), not net displacement. The particle reverses direction inside the interval, so we must split the motion. Step 1 — Position and velocity (using α=1, β=6, γ=5): s(t)=t²-6t+5, v(t)=ds/dt=2t-6. v=0 when 2t-6=0 → t=3 s, so the particle reverses at t=3 s. Step 2 — Find positions to get distances on each leg. s(0)=5 m, s(3)=9-18+5=-4 m, s(6)=36-36+5=5 m. Distance on [0,3]: |s(3)-s(0)|=|-4-5|=9 m. Distance on [3,6]: |s(6)-s(3)|=|5-(-4)|=9 m. Total distance = 9+9 = 18 m. Step 3 — Average speed. Average speed = total distance / total time = 18/6 = 3 ms⁻¹. This matches option C. Trap: the net displacement is s(6)-s(0)=0, so the average VELOCITY is 0 (option D). The question asks for average SPEED, which is 3 ms⁻¹ — distance must include both 9 m legs.