Question

(ii) Brakes were applied on a train running at 60km/h to stop in 5 minutes on a coming railway
station. Calculate distance travelled by the train before it comes to rest.
(5)

Answer 1

Answer :

➥ The distance travelled by the train = 2523.18 m

Given :

➤ Initial velocity of a train (u) = 60km/h

➤ Final velocity of a train (v) = 0 m/s

➤ Time taken by a train (t) = 5 min

To Find :

➤ Distance travelled by the train (s) = ?

Required Solution :

To find the Distance travelled by the train, at first we need to convert the units of Initial velocity km/h to m/s and we also need to convert time taken by the train minute to second, after converting the units we will find the Acceleration of the train, after all this we will find the Distance travelled by the train on the basis of conditions given above.

◈ Initial velocity = 60 km/h

→ Initial velocity = 60 × 5/18

→ Initial velocity = 300/18

→ Initial velocity = 16.66 m/s

◈ Time taken = 5 min

→ Time taken = 5 × 60

→ Time taken = 300 sec

We can find Acceleration of the train by using the first equation of motion which says v = u + at.

Here,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• a is the Acceleration in m/s.
• t is the Time taken in second.

So, let's calculate Acceleration (a) !

From first equation of motion

⇒ v = u + at

⇒ 0 = 16.66 + a × 300

⇒ 0 = 16.66 + 300a

⇒ 0 - 16.66 = 300a

⇒ -16.66 = 300a

⇒ 16.66/300 = a

⇒ -0.055 = a

⇒ a = -0.055 m/s²

【NOTE : Minus/Negative sign show retardation.】

Now we have three elements that used in formula, Initial velocity, Final velocity, Acceleration of the train.

• Initial velocity of a train = 16.66 m/s
• Final velocity of a train = 0 m/s
• Acceleration of a train -0.055 m/s²

And we need to find Distance travelled by the train$.$

We can find Distance travelled by the train by using the third equation of motion which says v² = u² + 2as.

Here,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• A is the Acceleration in m/s².
• a is the Distance in m.

So let's find Distance (s) !

⇛ v² = u² +2as

⇛ 0² = 16.66² + 2 × (-0.055) × s

⇛ 0 = -277.55 + (-0.11) × a

⇛ 0 - 277.55 = 0.11 × s

⇛ -277.55 = 0.11s

⇛ -277.55/0.11 = s

⇛ 266.55/0.11 = s

⇛ 2523.18 = s

⇛ s = 2523.18

║Hence, the Distance travelled by the train is 2523.18 m.║