The driver of a passenger trin moving at a speed of `90 km h^(-1)` observes a goods train at rest on the same track at a distance of 250 m and immediately applies the brakes. What should be the minimum retradation of the passenger train so as the avoid collision of the two traisn?

Explanation

The initial velocity of the passanger train `u=90 km h^(-1)`
`=90xx(5)/(18) m s^(-1)=25 m s^(-1)`
The final velocity of the passanger train ,` v=0 m s^(-1)`
The distane, tht the can be travelled by the passanger train in coming to rest position, s=250 m.
Let 't' be the time taken by thee passanger train to come to rest. Then, the total distance travelled by the train a= (Average velocity) (time)
`rArrS=((v+u)/(2))t`
`rArr=250=((0+25)/(2))t`
Let 'a' be the acceleration of the passanger train, then `a=(v-u)/(t)`
`=(0-25)/(20)=(-25)/(20)`
Hecnce, retardation of teh passneger train is `1.25 m s^(-2)`