A particle moves in a straight line. Its velocity-time graph shows a constant velocity of 15 m/s for the first 4 s , and an unknown constant negative velocity -v₀ (where v₀ > 0 ) for the next 6 s (from t = 4 s to t = 10 s ). If the ratio of the total distance covered to the magnitude of displacement in the 10 s interval is 4:1 , the value of v₀ is

Step-by-step solution

The area under the velocity-time graph gives the displacement, and the sum of the absolute areas gives the distance. For the first 4 s , the velocity is constant at 15 m/s . Area A₁ = 15 4 = 60 m . For the next 6 s (from t = 4 s to t = 10 s ), the velocity is -v₀ . Area A₂ = -v₀ 6 = -6v₀ m . Total distance covered = |A₁| + |A₂| = 60 + 6v₀ . Magnitude of displacement = |A₁ + A₂| = |60 - 6v₀| . Assuming the net displacement is positive (which is consistent with the options provided), we have: 60 + 6v₀ 60 - 6v₀ = 4 So