The velocity-time ( v - t ) graph of a body projected vertically upwards from the ground is a straight line. The graph intersects the velocity axis at 20 m/s and the time axis at 2 s . What is the ratio of the total distance covered to the magnitude of displacement of the body at t = 3 s ?

Step-by-step solution

From the given information, the velocity at t = 0 is u = 20 m/s , and the velocity becomes zero at t = 2 s . The acceleration is the slope of the v - t graph: a = 0 - 20 2 - 0 = -10 m/s ^2 At t = 3 s , the velocity is: v = u + at = 20 - 10(3) = -10 m/s The displacement is the net area under the v - t graph. The graph forms two triangles with the time axis: Area from t = 0 to t = 2 s (above the axis): A₁ = 1 2 2 20 = 20 m Area from t = 2 s to t = 3 s (below the axis): A₂ = 1 2 1 (-10) = -5 m Magnitude of displacemen