Question

2. A train is travelling at a speed of 100 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.

Answer 1

Given :

▪ Initial speed of train = 100kmph

▪ Final speed of train = zero

▪ Retardation = -0.5m/s²

To Find :

▪ Distance covered by train before it is brought to rest.

Concept :

→ This question is completely based on the concept of stopping distance.

→ Since, acceleration has said to be uniform throughout the motion, we can easily apply equation of kinematics to solve this type of questions.

Conversion :

↗ 1kmph = 5/18mps

↗ 100kmph = 100×5/18 = 27.78mps

Calculation :

→ v² - u² = 2aS

→ (0)²- (27.78)² = 2(-0.5)(S)

→ S = 771.72m

[Note : -ve sign of a shows retardation.]

Answer 2

Answer:

Given:

• A train is travelling at a speed of 100 km/hr-1.
• Brakes are applied so as to produce a uniform acceleration of -0.5 m s-2.

Find:

• Find how far the train will go before it is brought to rest.

Note:

• 1 km/hr = 1000 m
• 1 hr = 60 × 60 = 3600 sec
• 1000 m/3600 sec = 10/36 m/s
• 10/36 = 5/18 m/s [.•. 1 km/hr = 5/18 m/s]
• According to question, 100 km/hr is equal to (5/18 × 100) = 27.78 m/s.

know terms:

1. Km = Kilometer.
2. hr = Hour.
3. sec = Second.
4. m/s = Minute/Second.
5. m = ameter.
6. v = Final velocity.
7. u = Initial velocity.
8. a = Acceleration.
9. s = Displacement.
10. (-ve) = Retardation.

Using formula:

⇒ v² - u² = 2 as

Calculation:

⇒ (0)² - (27.78)² = 2 (-0.5) (s)

⇒ s = 771.72 m