Answer:
Explanation:
Given
average speed of train[tex](v_{avg})=216 kmph\approx 60 m/s[/tex]
Maximum acceleration=0.05g
Now centripetal acceleration is
[tex]a_c=\frac{v^2}{r}[/tex]
[tex]0.05\times 9.8=\frac{60^2}{r}[/tex]
r=7346.93 m
(b)Radius of curvature=900 m
therefore [tex]a_c=\frac{v^2}{r}[/tex]
[tex]v=\sqrt{a_cr}[/tex]
[tex]v=\sqrt{0.05\times 9.8\times 900}[/tex]
[tex]v=\sqrt{441}=21 m/s[/tex]