Two particles A and B move along a straight line. The velocity of particle A varies with time t as follows: it increases linearly from zero to 20 m/s in the first 2 s , remains constant at 20 m/s from t = 2 s to t = 6 s , and then decreases linearly to -20 m/s at t = 10 s . Particle B starts its motion at t = 0 with an initial velocity of 5 m/s and moves with a constant acceleration a . If both particles have the sam

Step-by-step solution

The displacement of particle A is the algebraic sum of the areas under its velocity-time graph. Area from t = 0 to 2 s (triangle): A₁ = 1 2 2 20 = 20 m Area from t = 2 to 6 s (rectangle): A₂ = 4 20 = 80 m From t = 6 to 10 s , the velocity decreases linearly from 20 m/s to -20 m/s . It crosses zero at t = 8 s . Area from t = 6 to 8 s (positive triangle): A₃ = 1 2 2 20 = 20 m Area from t = 8 to 10 s (negative triangle): A₄ = 1 2 2 (-20) = -20 m Total displacement of particle A at t = 10 s : S_A = 20 + 80 + 20 - 20 =