The velocity-time graph of a particle moving along a straight line is a single straight line. At t = 0 , its velocity is 40 m/s . The velocity becomes zero at t = 8 s . The difference between the average speed and the magnitude of the average velocity of the particle over the time interval from t = 0 to t = 10 s is

Step-by-step solution

The velocity-time graph is a straight line, which implies constant acceleration. Acceleration a = v_f - v_i t = 0 - 40 8 - 0 = -5 m/s ^2 Velocity at t = 10 s is v = u + at = 40 + (-5)(10) = -10 m/s The displacement is the signed area under the velocity-time graph. Area from t = 0 to t = 8 s (triangle above axis): A₁ = 1 2 8 40 = 160 m Area from t = 8 s to t = 10 s (triangle below axis): A₂ = 1 2 (10 - 8) (-10) = -10 m Total distance = |A₁| + |A₂| = 160 + 10 = 170 m Total displacement = A₁ + A₂ = 160 - 10 = 150 m Av