Initial speed u = 54 km h⁻¹, final speed v = 36 km h⁻¹, and time = 36 s = 0.01 h. Average speed = 45 km h⁻¹. Distance = 45 × 0.01 = 0.45 km = 450 m.
A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
Share
Given, initial speed u = 54 km h⁻¹, final speed v = 36 km h⁻¹, and time t = 36 s = 0.01 h. Since acceleration is constant, average speed = (u + v)/2 = (54 + 36)/2 = 45 km h⁻¹. Therefore, distance travelled = average speed × time = 45 × 0.01 = 0.45 km = 450 m.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/