Answer:
Let s be length of train, t be time taken by full train to cross the pole and a be uniform acceleration of train.
from,\[{{v}^{2}}-{{u}^{2}}=2as,\frac{{{80}^{2}}-{{60}^{2}}}{2a}=s\]
or \[s=\frac{6400-3600}{2a}=\frac{1400}{a}\] ….(1)
The middle part of the train covers half the total distance,\[s/2=\frac{700}{a}\]
From,\[{{v}^{2}}-{{u}^{2}}=2as,\,\,{{v}^{2}}-{{60}^{2}}=2a\times \frac{700}{a}=1400\]
\[{{v}^{2}}=1400+3600=\sqrt{5000}=70.7\,km/h.\]
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