# CBSE Class 11-science Answered

**A train starts from a station with a constant acceleration of 4 m/s2. Two seconds later another train passes the same station on the same line and in the same direction with a constant velocity. Find the velocity if the collision is first averted**

First train has started from station with acceleration a = 4 m/s^{2} . 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 t^{2 }= (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 , d_{2} = ( v t )

Distance travelled by first train during this time t is , d_{1} = u t + (1/2) a t^{2}

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, d_{1} = 8 t^{ }+ (1/2) × 4 × t^{2} = 2 t^{2} + 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 = ( d_{1} + 8 ) - d_{2} = ( 2 t^{2} + 8 t + 8 ) - ( v t )

S = 2 t^{2} - ( 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 = 2t^{2} - ( 4 t ) t + 8 = 8 - 2 t^{2}

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.