Asked by Victoria
The Speed Of A Train Is Reduced From 80km/h to 40km/h In A Distance Of 500m On Applying The Brakes Calculate (I) How Much Further Will The Train Travel Before Coming To Rest (II) Assuming The Retardation Remain Constant,how Long Will It Take To Bring The Train To Rest After The Application Of The Brakes
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Henry = Wt. of
V = 40km/h = 40,000m/3600s = 11.1 m/s.
Vo = 80km/h = 80,000m/3600s = 22.2 m/s.
1. V^2 = Vo^2 + 2a*d = (11.1)^2.
(22.2)^2 + 2a*500 = (11.1)^2,
1000a = (11.1)^2-(22.2)^2 = -369.63, a = -0.370 m/s^2.
(22.2)^2 + 2*(-0.370)*d = 0.
-0.740d = -(22.2)^2, d = 666 m. further.
2. V = Vo + a+t = 0.
22.2 - 0.370*t = 0, t = 60 s.
Vo = 80km/h = 80,000m/3600s = 22.2 m/s.
1. V^2 = Vo^2 + 2a*d = (11.1)^2.
(22.2)^2 + 2a*500 = (11.1)^2,
1000a = (11.1)^2-(22.2)^2 = -369.63, a = -0.370 m/s^2.
(22.2)^2 + 2*(-0.370)*d = 0.
-0.740d = -(22.2)^2, d = 666 m. further.
2. V = Vo + a+t = 0.
22.2 - 0.370*t = 0, t = 60 s.
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