Question

The speed of a train is reduced from 60 km/h to 15 km/h while it travels a distance of 450 m. If the retardation is uniform, find how much further will it travel before coming to rest (in m)?

Answer 1

Answer

Here speed of a train is reduced from 60 km/h to 15 km/h , when it travels a distance of 450 m.

Initial velocity , u = 60 km/h = 16.67 m/s

Final velocity , v = 15 km/h =  4.167 m/s

Distance , s = 450 m

Now using 3rd equation of motion ,

• v² - u² = 2 a s

⇒ (4.167)² - (16.67)² = 2 a (450)

⇒ 17.36 - 277.88 = 900 a

⇒ 900 a = - 260.52

⇒ a = - 0.289 m/s²

Here -ve symbol of acceleration shows retardation.

Given that retardation is uniform . we need to calculate how far will it move before coming to rest .

Initial velocity , u = 15 km/h = 4.167 m/s

Final velocity , v = 0 m/s

Acceleration , a = - 0.289 m/s²

Now using 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ 0² - (4.167)² = 2(-0.289)s

⇒ s = (4.167)²/(2*0.289)

⇒ s = 30.04 m

So it travels 30.04 meters before coming to rest.