The velocity-time graph of a particle moving in a straight line consists of two triangular regions. From t = 0 to t = 4 s , the velocity is positive, forming a triangle with a peak velocity of v₀ . From t = 4 s to t = t₂ , the velocity is negative, forming another triangle with a peak velocity of -v₀ . If the ratio of the total distance covered to the magnitude of net displacement from t = 0 to t = t₂ is 3 : 1 , the

Step-by-step solution

The area under the velocity-time graph gives the displacement, and the sum of the absolute areas gives the total distance. Let A₁ be the area of the positive triangle (from t = 0 to t = 4 s ): A₁ = 1 2 base height = 1 2 4 v₀ = 2v₀ Let A₂ be the magnitude of the area of the negative triangle (from t = 4 s to t = t₂ ): A₂ = 1 2 (t₂ - 4) v₀ The total distance is D = A₁ + A₂ . The net displacement is S = A₁ - A₂ . Given that the ratio of distance to displacement is 3 : 1 , we have: A₁ + A₂ A₁ - A₂ = 3 Substituting the