Question

A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s2. Find how for the train will go before it is brought to rest.

Answer 1

Answer:

625 metres

Explanation:

u = 90 km / h = 90 × [1000 / (60 ×60)]  m / s = 25 m/s

v = 0 m/s (∵ the train stops)

a = -0.5 m/s²

Using Newton's third equation of motion,

v² - u² = 2as ⇒ s = (v² - u²)/ 2a

⇒ s = (0 - 25²) / 2 × - 0.5

⇒ s = -625 / 1.0 = -625 m

∵ distance cannot be negative,

∴ distance travelled = 625 metres

Answer 2

Answer:

625 meters.

Explanation:

u=90km/h

(90×1000/3600) m/s

= 25/ms.

u = 0m/s (when brakes are applied)

a = (-0.5) m/s²

u² = u²+2as (acc. to third equation of motion)

0 = (25)² + (2) × (0.5) s

= 625 - 1s

1s = 625 m

s = 625/1 m

Therefore,

s = 625 m