Question

Shatabdi express starts from rest and attains a velocity of 108km/h in 2min . If acceleration is uniform fild acceleration and displacement of the train..

Answer 1

Answer :

Acceleration of the train is 0.25m/s²

Displacement of the train is 1.8km

Explanation :

According to the Question

It is given that ,

• Initial velocity ,u = 0m/s
• Final velocity ,v = 108×5/18 = 30m/s
• Time taken ,t = 2min = 2*60 = 120s

we have to calculate the uniform acceleration of the train and displacement of the train .

Firstly we calculate the uniform acceleration of the train .

Using Kinematics Equation

• a = v-u/t

Substitute the value we get

➾ a = 30-0/120

➾ a = 30/120

➾ a = 1/4

➾ a = 0.25m/s²

• Hence, the acceleration of the train is 0.25m/s²

Now, calculating the displacement of the train .

Again using Kinematics Equation

• v² = u² + 2as

Puttng all the values we get

➾ 30² = 0² + 2×0.25 × s

➾ 900 = 0 + 0.50 × s

➾ 900 = 0.50×s

➾ s = 900/0.50

➾ s = 1800m

➾ s = 1.8km

• Hence, the displacement of the train is 1.8 kilometres.

Answer 2

Given :-

Shatabdi express starts from rest and attains a velocity of 108km/h in 2min

To Find :-

Acceleration

Displacement

Solution :-

We know that

⮆  1 km/h = 5/18 m/s

⮆  108 km/h = 108 × 5/18 m/s

⮆  108 km/h = 6 × 5 m/s

⮆  108 km/h = 30 m/s

⮆  1 min = 60 sec

⮆  2 min = 2 × 60 = 120 sec

Now,

◼ F i n d i n g A c c e l e r a t i o n :-

a = (v - u)/t

Here,

a = acceleration

v = final velocity

u = initial velocity

t = time

⮆  a = (30 - 0)/120

⮆  a = 30/120

⮆  a = 1/4 m/s²

◼ F i n d i n g D i s p l a c m e n t :-

s = ut + 1/2 at²

Here,

s = distance

u = initial velocity

t = time

a = acceleration

⮆  s = 0 × 120 + 1/2 × 1/4 × (120)²

⮆  s = 0 + 1/8 × 14400

⮆  s = 14400/8

⮆  s = 1800 m