Question

An observer stands on the platform at the front edge of the first bogie of a stationary train. the train starts moving with uniform acceleration and the first bogie takes 5 seconds to cross the observer. if all the bogies of the train are of equal length and the gap between them is negligible, the time taken by the tenth bogie to cross the observer is

(a) 1.07s
(b) 0.98s
(c) 0.91s
(d) 0.81s

Answer 1

The tenth bogie will take 1.07 seconds to cross the observer because the distance it needs to cover is the same as the first bogie, and its relative speed is also the same.

Option A is correct

The observer on the platform sees the tenth bogie move with a relative speed compared to the train itself. This relative speed is equal to the speed of the first bogie, as both are part of the moving train.

We can relate the relative speed (v), distance (d), and time (t) using the formula: d = v * t

The distance traveled by the first bogie in 5 seconds (d1) is also the distance the tenth bogie needs to cover. We know that

d1 = v * t1 and d2 = d1 (since both travel the same distance).

We then substitute for d2, we have: v * t2 = v * t1

Dividing both sides by v, we get: t2 = t1

In conclusion, the time taken for the tenth bogie to cross the observer (t2) is equal to the time taken by the first bogie (t1), which is 5 seconds.