Given, initial velocity u = 0, final velocity v = 24 m s⁻¹ and time t = 6 s.
Average acceleration = (24 − 0)/6 = 4 m s⁻².
Distance = ½(0 + 24) × 6 = 72 m.
A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
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The initial velocity of the car is u = 0 m s⁻¹, final velocity is v = 24 m s⁻¹ and time is t = 6 s.
Average acceleration = (v − u)/t = (24 − 0)/6 = 4 m s⁻².
Distance travelled = average velocity × time = [(u + v)/2] × t = [(0 + 24)/2] × 6 = 72 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/