A particle has initial velocity +25 m s⁻¹ and constant acceleration -4 m s⁻². The actual distance travelled during the 7th second is:
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A particle has initial velocity +25 m s⁻¹ and constant acceleration −4 m s⁻². The actual distance travelled during the 7th second is:

G1 Physics · Accepted Answer

The 7th second is from t=6 s to t=7 s. Velocity: v = 25−4t. The particle stops at 25−4t=0 ⇒ t=6.25 s, so the reversal occurs inside the 7th second. From t=6 to 6.25: v changes 1 → 0 m s⁻¹. Distance = (1+0)(0.25)/2 = 0.125 m. From t=6.25 to 7: speed changes 0 → 3 m s⁻¹. Distance = (0+3)(0.75)/2 = 1.125 m. Total distance = 1.25 m.