The velocity-time graph of a particle over a 40 s interval is shown in the figure. The graph consists of two triangular regions of equal base 20 s, one above and one below the time axis, each reaching a speed magnitude of 5 m·s⁻¹. What are, respectively, the total distance travelled by the particle and its average velocity over the 40 s interval?

Question

G1 Physics · Accepted Answer

The distance travelled equals the total area between the velocity-time graph and the time axis, taking every area as positive. Area of the positive triangular part: A₁ = (1 ÷ 2)(20)(5) = 50 m. Area of the negative triangular part contributes another 50 m to the distance. Therefore, total distance = 50+50 = 100 m. For displacement, the signed areas cancel: displacement = 50-50 = 0. Hence, average velocity = displacement ÷ total time = 0 ÷ 40 = 0 m·s⁻¹. Correct option: C.