From the v-t graph shown, the ratio of distance to displacement in 25 s of motion is:

Solution

Physics sir · 3 steps

Area under the v-t graph: keep the signs for displacement, drop them for distance.

1. Where it turns. From 15 s to 25 s the velocity falls in a straight line from 20 to -20 m s⁻¹. It reaches zero halfway, at t = 20 s. After that the body moves backwards.

   Interval	Shape	Area (m)
   0 to 5 s	triangle	1/2 × 5 × 10 = 25
   5 to 10 s	rectangle	5 × 10 = 50
   10 to 15 s	trapezium	1/2 × (10 + 20) × 5 = 75
   15 to 20 s	triangle above axis	1/2 × 5 × 20 = 50
   20 to 25 s	triangle below axis	-1/2 × 5 × 20 = -50

2. Distance and displacement. Distance adds every area as positive: 25 + 50 + 75 + 50 + 50 = 250 m. Displacement keeps the sign: 200 - 50 = 150 m.

   distance/displacement = 250/150 = 5/3

3. So. The ratio is 5/3, option (d).

The idea to keep. On a v-t graph, area below the time axis is backward motion. Count it as negative for displacement and as positive for distance. So distance is never less than displacement.

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