The speed of a train is reduced from 80km/h to 40km/h in a distance of 500m on applying the brake
80+at = 40
at = -40
80t + 1/2 at^2 = 1/2
80t - 1/2 * 40 * t = 1/2
60t = 1/2
t = 1/120
1/120 a = -40
a = -4800 km/hr^2
Correction:a = -4.8km/h^2.
To find the acceleration of the train, we can use the equation:
acceleration (a) = (final velocity - initial velocity) / time
Here, the final velocity (v) is 40 km/h, the initial velocity (u) is 80 km/h, and the distance (s) is 500 m. We need to convert the velocities from km/h to m/s since the distance is given in meters.
Step 1: Convert velocities to m/s:
To convert km/h to m/s, divide the value by 3.6.
Final velocity (v) = 40 km/h = 40/3.6 m/s = 11.11 m/s
Initial velocity (u) = 80 km/h = 80/3.6 m/s = 22.22 m/s
Step 2: Calculate the acceleration:
acceleration (a) = (v - u) / t
Given that the distance (s) = 500 m, we can use the formula: s = ut + (1/2)at^2, where t is unknown.
Rewriting the formula: t = (v - u) / a
Substituting the values into the formula:
500 = 22.22 * t + (1/2) * a * t^2
We have two unknowns, t and a. To find them, we need additional information.
Please provide the time it took for the train to decelerate from 80 km/h to 40 km/h.
V^2 = Vo^2 + 2a*d = 40^2
80^2+2a*500 = 1600
641000a = 1600
1000a = -4800
a = - 4.8m/h^2.