Asked by ivy
A train braking with constant deceleration covers 1km in 20s, and a second kilometre in 30s. Find the Deceleration. Wgat further distance will it cover before coming to a stop. and how long will it take?
Answers
Answered by
Henry
1km = 1000m.
Vo = 1000m/20s = 50m/s.
Vf = 1000m/30s = 33.33m/s.
a = (Vf-Vo)/t = (33.33 - 50)/(30-20) =
-16.67 / 10 = -1.67m/s^2.
The negative sign meane deceleration.
Vf^2 = Vo^2 + 2ad = 0,
(33.33)^2 + 2*(-1.67)d = 0,
1111.1 -3.34d = 0,
-3.34d = -1111.1,
d = 332.7m.
Vf = Vo + at,
t = (Vf-Vo)/a,
t = (0 - 33.33) / -1.67 = 20s.
Vo = 1000m/20s = 50m/s.
Vf = 1000m/30s = 33.33m/s.
a = (Vf-Vo)/t = (33.33 - 50)/(30-20) =
-16.67 / 10 = -1.67m/s^2.
The negative sign meane deceleration.
Vf^2 = Vo^2 + 2ad = 0,
(33.33)^2 + 2*(-1.67)d = 0,
1111.1 -3.34d = 0,
-3.34d = -1111.1,
d = 332.7m.
Vf = Vo + at,
t = (Vf-Vo)/a,
t = (0 - 33.33) / -1.67 = 20s.
Answered by
Mahnoor
This is wrong
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