A person starts his journey from centre 'O' of the park and comes back to the same position following path OPQO as shown in the figure. The radius of path taken by the person is 200 m and he takes 3 min 58 sec to complete his journey. The average speed of the person is _____________ ms⁻¹. (take π = 3.14)

Solution

Physics sir · 4 steps

Average speed needs the total path length, so add up both radii and the quarter arc.

1. Read the path. The person goes from O to P along a radius, from P to Q along the circle, then from Q back to O along another radius. OP and OQ are at right angles, so the arc PQ is a quarter circle.

   d = r + (2π r)/4 + r = 200 + (3.14 × 200)/2 + 200 = 714 m

2. Total time. 3 min 58 s = 180 + 58 = 238 s.
3. Divide. Average speed = (total distance)/(total time) = 714/238 m s⁻¹.
4. So. 714 = 3 × 238, so the average speed is 3 m s⁻¹.

The common slip: 0. Dividing displacement by time gives 0, which is the average velocity. Average speed needs the 714 m of path actually walked.

The idea to keep. Average speed uses the full distance travelled, and average velocity uses displacement. A round trip back to the start has zero displacement but never zero distance.

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