Practice question

Given: A particle moves with an initial velocity of 4 i m/s and an acceleration of (2 i - 3 j) m/s² . What is its speed after 3 s ? These values define the system as per NCERT data. Formula: Velocity v = v_0 + a t. This is standard NCERT relation. Substitution & Calculation: Given: v_0 = 4 i, a = 2 i - 3 j, t = 3 s . v = 4 i + (2 i - 3 j) × 3 = 4 i + 6 i - 9 j = 10 i - 9 j m/s . Speed v = sqrt10²+ (-9)² = sqrt100 + 81 = sqrt181 approx 13.45 m/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.