Question

A train runs between two stations with acceleration 1 m/s² and then with retardation 2 m/s². If distance between stations is 120 Km., What is minimum time and maximum speed attained in journey?

Answer 1

Now, the train will start from rest from station A, accelerate with 1 m/s² to attain a top speed of say 'v'. Now, after that it will decelerate with 2 m/s² and eventually halt at station B.

So,

During acceleration

v₁ = u₁ + a₁t₁

here,

u₁ = 0

a₁ = 1 m/s²

thus,

v₁ = 0 + 1.t₁

or

v₁ = t₁

and also

v₁² - u₁² = 2a₁s₁

so,

v₁ = [2a₁s₁]^1/2

or [as a₂ = 1 m/s²]

v₁ = [2s₁]^1/2

so, time taken to accelerate [as v₁ = t₁]

t₁ = [2s₁]^1/2

or distance travelled in this case

s₁ = t₁² / 2 ..................................(1)

and

During deceleration

v₂ = u₂ + a₂t₂

here

v₂ = 0

u₂ = v₁

a₂ = -2 m/s²

so,

-v₁ = -2t₂

or

v₁ = 2t₂ so,

t₂ = v₁/ 2 .....................................(2)

also

v₂² - u₂² = 2a₂s₂

so, [substututing values]

v₁ = [2a₂s₂]^1/2

or [as a₂ = 2 m/s²]

v₁ = [4s₂]^1/2

so, time taken to decelerate [as v₁ = 2t₂]

2t₂ = [4s₂]^1/2

or distance travelled in this case

s₂ = t₂² ..................................(3)

..

now, we know that

s₁ + s₂ = 120km

so,

from (1) and (3),

( t₁² / 2) + t₂² = 120

also

as

v₁ = t₁

and also

v₁ = 2t₂ ; t₂ = v₁ / 2

we get,

( v₁² / 2) + (v₁ / 2)² = 120 x 10³ m

(3/4)v₁² = 120000

v₁² = 160000

thus, maximum speed will be

v₁ = 400 m/s

so,

minimum time will be

T = t₁ + t₂ = v₁ + (v₁ / 2)

= 400 + (400/2)

thus,

T = 600s