if a train at 72kmphr is to be brought to rest in a distance of 200 m then its retardation is A 20m/s^2 b 10m/s^2 c 2m/s^2 d 1m/s^2
72/3.6 = 20 meters/second
(helpful to remember 1000 meters /3600 seconds = 1/3.6)
0 = 20 + a t
t = -20/a
x = Xi + Vi t + (1/2) a t^2
200 = 0 + 20 t + .5 a t^2
200 = 0 -400/a + .5 a (400/a^2)
200 a = -400 + 200
a = -1
so d, retardation of 1 m/s^2
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alternatively
average speed during stop = 10 m/s^2
200 = 10 t
t = 20
a = change in velocity/ time = -20/20 = -1
To calculate the retardation of a train, we can use the formula:
Retardation (a) = (Final velocity^2 - Initial velocity^2) / (2 * Distance)
Given:
Initial velocity (u) = 72 km/hr = (72 * 1000) / 3600 m/s = 20 m/s (approx.)
Final velocity (v) = 0 (since the train comes to rest)
Distance (s) = 200 m
Now, let's substitute these values into the formula:
Retardation (a) = (0 - 20^2) / (2 * 200)
= (-400) / 400
= -1 m/s^2
The negative sign indicates that the train is decelerating or slowing down. Therefore, the correct answer is option (d) 1 m/s^2.