Vo = 36km/h = 36,000m/3600s = 10 m/s.
V = 9km/h = 9000m/3600s = 2.5 m/s.
a = (V^2-Vo^2)/2d
a = (2.5^2-10^2)/300 = -0.3125 m/s^2.
d = (V^2-Vo^2)/2a
d = (0-2.5^2)/-0.6250 = 10 m. Further.
The speed of the train is reduced from 36 km per hour to 9 km per hour whilst it travels a distance of 150 metres,if the retardation be uniform,find how much further it will travel before coming to rest
3 answers
Or:
d = -(10^2)/-0.6250 = 160 m.
160 - 150 = 10 m. Further.
d = -(10^2)/-0.6250 = 160 m.
160 - 150 = 10 m. Further.
Vo = 36km/h = 36,000m/3600s = 10 m/s.
V = 9km/h = 9000m/3600s = 2.5 m/s.
a = (V^2-Vo^2)/2d
a = (2.5^2-10^2)/300 = -0.3125 m/s^2.
d = (V^2-Vo^2)/2a
d = (0-2.5^2)/-0.6250 = 10 m.
d = -(10^2)/-0.6250 = 160 m.
160 - 150 = 10 m. Further.
V = 9km/h = 9000m/3600s = 2.5 m/s.
a = (V^2-Vo^2)/2d
a = (2.5^2-10^2)/300 = -0.3125 m/s^2.
d = (V^2-Vo^2)/2a
d = (0-2.5^2)/-0.6250 = 10 m.
d = -(10^2)/-0.6250 = 160 m.
160 - 150 = 10 m. Further.
1 answer