Two train the moving on the same track. the velocity of first train and 72 km/h and the second is 36 km/h. If the 1st train behind the 2nd train, then the minimum distance between the trains such that they do not collide ( retardation of brakes of 1st train is 0.05 m/s^2)

1) 1000m

2) 500m

3) 100m

4) 3000m

Asked by kamlesh.chahal.2003 | 20th Jul, 2019, 06:09: PM

Expert Answer:

speed of first train = 72 km/hr = 72×(5/18) m/s = 20 m/s
speed of second train = 36 km/hr = 36×(5/18) = 10 m/s
 
since the first train is behind second train, it has to reduce its speed 20 m/s to 10 m/s or
less than 10 m/s in order not to collide with the second train.
 
distance S required to reduce the speed is obtained from the equation, v2 = u2 - (2 a S)  ,
where v is final speed , u is initial speed and a is retardation
 
Hence distance S required for first train to reduce the speed = ( 202 - 102 )/(2×0.05) = 3000 m
 
time taken to reduce the speed = (v-u)/a  = (20-10)/0.05 = 200 s 
 
distance travelled by second train in 200 s = 10×200 = 2000 m
 
Hence required distance to be kept between first and second train to prevent collision = ( 3000 - 2000 ) m= 1000 m

Answered by Thiyagarajan K | 21st Jul, 2019, 11:42: AM

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