NCERT Solutions for Class 9 Science Chapter 8
Motion
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Intext Questions
Page No-110
Questions No-2
A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.
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Here, we have,
Initial velocity, u = 90 km/h = 90×1000/3600 ms⁻¹=25 ms⁻¹
Final velocity, v = 0 m/s
Acceleration, a = – 0.5 m/s²
Distance travelled = ?
Using, v2 = u2 + 2as
s=v²−u²/2a=0²−25²/2(−0.5) = 625 m
Therefore, train will go 625 m before it brought to rest.
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