The velocity-time relationship of a particle moving along a straight line is described as follows: at t = 0 , its velocity is 10 m/s . The velocity decreases linearly with time, reaching -10 m/s at t = 4 s . From t = 4 s to t = 6 s , the particle moves with a constant velocity of -10 m/s . The ratio of the total distance covered by the particle to the magnitude of its displacement at t = 6 s is:

Step-by-step solution

The motion can be divided into three time intervals based on the velocity-time graph. From t = 0 to t = 4 s , the velocity decreases linearly from 10 m/s to -10 m/s . The velocity becomes zero at the midpoint, t = 2 s . Area A₁ (from t = 0 to t = 2 s ) = 1 2 2 10 = 10 m Area A₂ (from t = 2 s to t = 4 s ) = 1 2 2 (-10) = -10 m From t = 4 s to t = 6 s , the velocity is constant at -10 m/s . Area A₃ (from t = 4 s to t = 6 s ) = 2 (-10) = -20 m The displacement is the algebraic sum of the areas: Displacement = A₁ + A₂