FINAL VELOCITY FROM ACCELERATION

Constant acceleration means velocity increases at a steady rate every second. Starting at v₀ = 5 m/s with a = 2 m/s², each second adds 2 m/s to the speed. After 12 seconds, the total gain is 2 × 12 = 24 m/s, so v_f = 5 + 24 = 29 m/s. That is the first kinematic equation: v_f = v₀ + a·t. For the distance, we need the second equation: d = v₀·t + ½·a·t². The first term, v₀·t = 5 × 12 = 60 m, is the distance the train would have covered if it had stayed at its initial speed. The second term, ½ × 2 × 144 = 144 m, is the extra distance contributed by the acceleration. Together: d = 60 + 144 = 204 m. A common error is forgetting the ½ factor — which comes from integrating a constant acceleration — and writing a·t² = 288 m instead. The factor of ½ is not optional; it is the direct result of the area under a triangular velocity-time graph.