Question

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

Answer 1

Answer:

625 metres

Explanation:

Given:

Initial velocity = u = 90 km/h

$90 km/h = 90 \times \frac{5}{18} = 25 m/s$

Acceleration = a = -0.5 m/s²

Final velocity = v = 0 m/s (As the train would finally come to rest)

To find:

Distance covered by the train before coming to rest (s)

We can use the third equation of motion which says:

v²-u²=2as

0²-25²=2×-0.5×s

0-625= -1s

-625= -1s

s = 625 metres

The train will cover a distance of 625 metres before coming to rest position

Answer 2

Train will cover 625 m before it's brought to rest.

Solution

Initial velocity of train = 90 km/h

u = 90 × (5/18)

∴ u = 25 m/s

Acceleration (a) = - 0.5 m/s²

Final velocity of train = 0 m/s [∵ The train is brought to rest]

☢ Using 1st equation of motion :

● v = u + at

⇒ 0 = 25 + (-0.5)t

⇒ 0 = 25 - 0.5t

⇒ 0.5t = 25

⇒ t = 25/0.5

⇒ t = 50 seconds.

∴ Time taken to come to rest = 50 seconds

☢ Using 2nd equation of motion :

● s = ut + ½ at²

⇒ s = (25 × 50) + ½ × (-0.5) × 50²

⇒ s = 1250 + ½ × (-0.5) × 2500

⇒ s = 1250 - (0.5 × 1250)

⇒ s = 1250 - 625

⇒ s = 625 m

∴ Distance covered by train = 625 m (Ans.)

___________________________

Shortcut :

We have u = 25 m/s

v = 0 m/s [As it comes to rest]

a = -0.5 m/s²

☢ Using 3rd equation of motion :

● v² = u² + 2as

⇒ 0² = (25)² + 2s × (-0.5)

⇒ 0 = 625 - s

⇒ s = 625 m

∴ Distance covered by train before coming to rest = 625 m (Ans.)