A train starts from rest from a station with acceleration $0.2$ sec on a striaght track and then comes to rest after attianing maximum speed on another station due to retardation if total time spent is half is half an hour then, distance between two stations is:

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Solution

If A is the acceleration of the train and B is the retardation of the train, then the total distance traveled is given by $\frac{A B}{A + B} \times \frac{T^{2}}{2}$
Here $A = 0.2 m / s^{2}$ and $B = 0.4 m / s^{2}$ and $T = \frac{1}{2} h r = 1800 s e c$
Therefore $x = \frac{0.2 \times 0.4}{0.2 + 0.4} \times 1800 \times \frac{1800}{2} = 216000 m$
$= 216 k m$

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A train starts from rest from a station with acceleration 0.2 sec on a striaght track and then comes to rest after attianing maximum speed on another station due to retardation if total time spent is half is half an hour then, distance between two stations is:

If A is the acceleration of the train and B is the retardation of the train, then the total distance traveled is given by ABA+B×T22Here A=0.2m/s2 and B=0.4m/s2 and T=12hr=1800secTherefore x=0.2×0.40.2+0.4×1800×18002=216000m =216km

A train starts from rest from a station with acceleration $0.2$ sec on a striaght track and then comes to rest after attianing maximum speed on another station due to retardation if total time spent is half is half an hour then, distance between two stations is: