Question

a train is travellung at the speed of 90k/h breaks are applied so as to produce a uniform motion acceleration of 0.5 find hoe far the train will go before it will brought to rest ​

Answer 1

Given:-

• Initial velocity , u = 90km/h

• Final velocity ,v = 0m/s

• Acceleration ,a = 0.5m/s²

To Find:-

• Distance covered by the train , s

Solution:-

Conversion of unit

90km/h = 90×5/18 = 25m/s

By using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = 25² + 2×0.5×s

→ 0 = 625 + 1×s

→ s = 625/1

→ s = 625m

∴ The distance covered by the train is 625 Metre.

Answer 2

Explanation:

Initial velocity ,u = 10m/s

Acceleration ,a = 8m/s²

Final velocity ,v = 40m/s

So By using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 40² = 10² +2×8×s

→ 1600 = 100 + 16×s

→ 1600-100 = 16×s

→ 1500 = 16×s

→ s = 1500/16

→ s = 93.75m

The length of Runway 93.80m