AIIMS 2002 Physics Equations of Motion MCQ Question

Initial velocity (u) = 0, Acceleration (a) = 2 m/s^2 and time during acceleration (t1) = 10 sec. Time during constant velocity (t2) = 30 sec and retardation (r) = -4 m/s^2 (-ve sign due to retardation). Distance covered by the particle during acceleration, s1 = ut1 + 1/2 at1^2 = (0×10) + 1/2 × 2 × (10)^2 = 100 m. And velocity of the particle at the end of acceleration, v = u + at1 = 0 + (2×10) = 20 m/s. Therefore distance covered by the particle during constant velocity (s2) = v × t2 = 20 × 30 = 600 m. Relation for the distance covered by the particle during retardation (s3) is v^2 = u^2 + 2as3 or, (0)^2 = (20)^2 + 2 × (-4) × s3 = 400 - 8s3 or, s3 = 400/8 = 50 m. Therefore total distance covered by the particle.