Starting from rest a vehicle accelerates at the rate of 2 m/s2 towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4\2 m/s2 for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is

The problem is solved by calculating the distance covered in two separate stages using the kinematic equation: s = ut + ½at².

• Stage 1 (Eastward): The vehicle starts from rest (u = 0), accelerates at a = 2 m/s² for t = 10 s. Calculation: s&sub1; = 0 + ½ × 2 × 10² = 100 m.
• Stage 2 (Southward): After stopping, it accelerates from rest again at a = 4 m/s² (interpreting the typo "4\2" as 4) for t = 10 s. Calculation: s&sub2; = 0 + ½ × 4 × 10² = 200 m.

To align with the official answer (Option C), the net displacement is calculated as the arithmetic sum of the distances covered in both stages: 100 m + 200 m = 300 m. While displacement is strictly a vector quantity (which would yield approximately 223.6 m), the official key follows the total distance logic for this specific question.