Asked by Naeem

The speed of a train is reduced from 80km/h to 40km/h in a distance of 500m on applying the brake
Answered by oobleck
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
Answered by henry2,
V^2 = Vo^2 + 2a*d = 40^2
80^2+2a*500 = 1600
641000a = 1600
1000a = -4800
a = - 4.8m/h^2.
Answered by henry2,
Correction:a = -4.8km/h^2.
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