During acceleration, distance = ½ × 5 × 20 = 50 m. At constant speed, distance = 20 × 10 = 200 m. During braking, distance = ½ × 6 × 20 = 60 m. Total distance = 310 m.
A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
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The car first accelerates from 0 to 20 m s⁻¹ in 5 s. Distance travelled = ½(u + v)t = ½(0 + 20) × 5 = 50 m. It then travels at 20 m s⁻¹ for 10 s: distance = 20 × 10 = 200 m. During braking, distance = ½(20 + 0) × 6 = 60 m. Therefore, total distance = 50 + 200 + 60 = 310 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/