Practice question

Given: A particle starts with velocity 6 j m/s and accelerates at (2 i + 4 j) m/s² . What is its speed after 2 s ? These values define the system as per NCERT data. Formula: Velocity v = v_0 + a t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 6 j, a = 2 i + 4 j, t = 2 s . v = 6 j + (2 i + 4 j) × 2 = 4 i + (6 + 8) j = 4 i + 14 j m/s . Speed v = sqrt4² + 14² = sqrt16 + 196 = sqrt212 approx 14.56 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.