Answer:
Time taken = 10400 s
Explanation:
Given:
Initial speed of the train, [tex]u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}[/tex]
Final speed of the train, [tex]v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}[/tex]
Displacement of the train, [tex]S=234\textrm{ km}=234\times 1000=234000\textrm{ m}[/tex]
Using Newton's equation of motion,
[tex] v - u = at\\a=\frac{v-u}{t}[/tex]
Now, using Newton's equation of motion for displacement,
[tex]v^{2}-u^{2}=2aS[/tex]
Now, plug in the value of [tex]a=\frac{v-u}{t}[/tex] in the above equation. This gives,
[tex]v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}[/tex]
Now, plug in 234000 m for [tex]S[/tex], 25 m/s for [tex]v[/tex] and 20 m/s for [tex]u[/tex]. Solve for [tex]t[/tex].
[tex]t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}[/tex]
Therefore, the time taken by the train is 10400 s.
