Question

Consider a train which can accelerate with an acceleration of 20cm/s2 and slow down with deceleration of 100cm/s².Find the minimum time for the train to travel between the stations 2 km apart.

Answer 1

Explanation:

let , upto X metres the train will accelerate & then it'll start decelerate for next-left (2000–x) metres

(2Km=2000m)

so consider for first case,

V'² = 0 + 0.2x

or, V'²= 0.2x ---------- eqn1

for second case,

0= V'²– 1(2000–x)

or, 2000= 0.2x + x ( see above)

or, x= 1666.67

now , put the value of x to find V' using eqn1

then find the t using formula ,

V'= 0+ 0.2t

then find the value of T , using formula,

0 = V' – 1T

or, V'= T

last of all, calculate t+T

it's the answer.

Answer 2

Answer:

acceleration of train , a = 20cm/s² = 0.2 m/s²

deceleration , d = - 100cm/s² =- 1 m/s²

Let the train start with acceleration for xkm or, 1000x m distances and deceleration for (2700 - 1000x) m [ I just change (2.7 - x) km into metre.]

so, equation of distance travelled by train in 1st case :- when we assume train start from rest then, intial velocity, u = 0

use formula, v² = u² + 2aS

v² = 0 + 2 × 0.2 × 1000x = 400x ----(1)

equation of distance travelled by train in 2nd case :- intial velocity in this case = final velocity of 1st case = v.finally train will be rest .

so, final velocity in this case = 0

deceleration = d and distance is S' = (2700-1000x) m

so, 0 = v² + 2(-1) × (2700-1000x)

v² = 5400 - 2000x -----(2)

solve equations (1) and (2),

5400 - 2000x = 400x

=> 5400 = 2400x

=> x = 54/24 = 9/4 km

so, v² = 400 × 9/4 = 900

v = 30 m/s

now, use formula v = u + at

30 = 0 + 0.2t

t = 150 sec

again, time taken in 2nd case

0 = 30 + (-1)t => t = 30 sec

hence, total time taken = 180 sec