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 :

Solution

Physics sir · 3 steps

Distance is the area under the v–t graph, and the part below the axis counts as positive too.

1. Area gives distance. On a v–t graph, the area between the line and the time axis is the displacement for that interval. For distance, take every area as positive, including the part below the axis after S.
2. Check point S. From B (3, 4) to C (5, -2) the slope is -6/2 = -3 m s⁻². So v falls from 4 to 0 in 4/3 s, which puts S at t = 13/3 ≈ 4.333 s, as given.

   Interval	Shape	Area (m)
   0 to 2 s	triangle, base 2, height 4	1/2 × 2 × 4 = 4
   2 to 3 s	rectangle, 1 by 4	1 × 4 = 4
   3 to 13/3 s	triangle, base 4/3, height 4	1/2 × 4/3 × 4 = 8/3
   13/3 to 6 s	triangle below axis, base 5/3, height 2	1/2 × 5/3 × 2 = 5/3

   Distance = 4 + 4 + 8/3 + 5/3 = 37/3 m

3. So. The body covers 37/3 m in 6 s, which is option (d). Subtracting the last area instead would give the displacement, 9 m, which is not what is asked.

The idea to keep. For distance from a v–t graph, split the graph where it crosses the time axis and add the magnitudes of all areas. Subtracting the areas below the axis gives displacement instead.

Answer from the NTA final answer key. Solution by PrepWiser.