If train travelling 72kmph is to be brought to rest distance of 200m then its retardation should be
Vo = 72,000m/3600s = 20 m/s.
V^2 = Vo^2 + 2a*d = 0,
20^2 + 2a*200 = 0,
a = -1.0 m/s^2.
72 km/h * 1000 m/km * 1 h/3600 s = Vi
average speed = Vi/2
time to stop = t = 200 / (Vi/2)
200 = (1/2) a t^2
a= 400 /(t^2) = magnitude of deacceleration in m/s^2
or
Vi = a t
a = Vi/t
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To find the retardation (deceleration) of the train, we can use the equations of motion.
The equation linking distance (d), initial velocity (u), final velocity (v), acceleration (a), and time (t) is:
d = ut + (1/2)at^2
In this case, we want to bring the train to rest from an initial velocity of 72 km/hr (which we need to convert to m/s) over a distance of 200 m. We can assume the final velocity (v) will be zero since the train comes to rest.
Converting 72 km/hr to m/s:
72 km/hr = 72 * (1000m / 3600s) = 20 m/s
Plugging the known values into the equation:
200 = (20 * t) + (1/2) * a * t^2
At this point, we have one equation with two unknowns (a and t). To solve for the retardation (a), we need another equation relating the two variables.
The equation linking initial velocity (u), final velocity (v), acceleration (a), and time (t) is:
v = u + at
Again, since we want the final velocity to be zero, we can rewrite the equation as:
0 = 20 + at
Rearranging this equation, we get:
t = -20/a
Now we can substitute this value of t back into the first equation:
200 = (20 * (-20/a)) + (1/2) * a * (-20/a)^2
Simplifying further, we have:
200 = -400/a + 200/a^2
Multiplying through by a^2, we get:
200a^2 = -400a + 200
Rearranging and simplifying, we have:
200a^2 + 400a - 200 = 0
Now we have a quadratic equation which we can solve using the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 200, b = 400, and c = -200. Plugging in these values, we can solve for a.
Solving the equation gives us two possible values for a:
a ≈ 0.682 m/s^2 (approximate value)
a ≈ -1.182 m/s^2 (approximate value)
Since retardation refers to a negative acceleration (deceleration), the appropriate value for the retardation of the train would be approximately 1.182 m/s^2.