At a distance L= 400m away from the signal light,brakes are applied to a locomotive moving with a velocity, u = 54 km/h.Determine the position of rest of the locomotive relative to the signal light after 1 min of the application of the brakes if its acceleration a = - 0.3 m/`s^(2)`

Explanation

Since the locomotive moves with a constant deceleration after the application of brakes, it will come to rest in .t. sec . We know ,
v = u + at
Here u = 54 km/h = 54 x 5/18 = 15 m/s
Let v = 0 at time .t. and given
a = - 0.3 m/`S^(2)`
From v = u + at we get t = `(v - u)/(a)`
We get , t `= (-15)/(-0.3) = 50 s `
During which it will cover a distance
s = - `(u^(2))/(2a)`
`= - (15^(2))/(2 xx (-0.3))`
= `(225)/(0.6)`
= 375 m
Thus in 1 min after the application of brakes the locomotive will be at a distance l = L – s = 400 – 375 = 25 m from the signal light.