The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is :
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Distance = total area between the v-t curve and the time axis, taking all areas as positive (the velocity is negative between S and D).
Split the motion into the part with v \(\ge\) 0 (0 to 4.333 s) and the part with v < 0 (4.333 s to 6 s). Forward part: the area is a trapezium with the long base 13/3 s (from t = 0 to S), the flat top 1 s (A to B) and height 4 m/s: area = \(\frac{1}{2}\left(\frac{13}{3}+1\right)\times 4\) = \(\frac{1}{2}\times\frac{16}{3}\times 4\) = \(\frac{32}{3}\) m. Backward part: a triangle with base 6 - 13/3 = 5/3 s and depth 2 m/s: area = \(\frac{1}{2}\times\frac{5}{3}\times 2\) = \(\frac{5}{3}\) m. Total distance = \(\frac{32}{3}+\frac{5}{3}\) = \(\frac{37}{3}\) m \(\approx\) 12.33 m (the displacement would be \(\frac{32}{3}-\frac{5}{3}\) = 9 m, but distance counts the backward part as positive). Answer: (4) \(\frac{37}{3}\) m
