CBSE Class 11-science Answered

First train has started from station with acceleration a = 4 m/s² . After 2 s , second train reached station that moves with constant speed v .

Distance travelled by first train in  2 s is

d_o = (1/2 ) a t² = (1/2) × 4 × 2 × 2 = 8 m

Let us find out the time t and speed v of second train so that second train reaches the point of first train .

Then speed less than this speed v is the required speed of second train to prevent collision .

Let t be the time of second train starting from station .

Then distance travelled by second train is , d₂ =  ( v t )

Distance travelled by first train during this time t is ,  d₁ = u t + (1/2) a t²

Speed u in above expression is speed of first train  2 seconds after it has started from station from rest.

u = a × t = 4 × 2 =8 m/s

Hence,  d₁ = 8 t + (1/2) × 4 × t² = 2 t² + 8 t

when the second train is at station , distance between first and secon train is 8 m .

Hence difference S between first and second train after t seconds is

S = ( d₁ + 8 ) - d₂ = ( 2 t² + 8 t + 8 ) - ( v t )

S = 2 t² - ( v - 8 ) t + 8  ......................... (1)

To get minimum S , we differentiate above expression and equate it to zero

dS/dt = 4t - (v-8) = 0

Hence S will be minimum  when  (v-8) = 4 t  ...................(2)

If we substitute (v-8) = 4t  in eqn.(1)  , then we get

S = 2t² - ( 4 t ) t + 8 =  8 - 2 t²

If S = 0 , then from above expression , we get t = 2

By substitutting t = 2 in eqn.(2) , we get v = 16 m/s

Hence , if second train travels with constant speed v = 16 m/s , then second train collides with first train in 2 s .

If second train travels with constant speed v < 16 m/s , then collision is avoided.