a train 100 m long and stationary is given the all clear signal 80 m ahead of it . the train then accelerates uniformly at 0.4 m/s² . find the time taken for the engine driver at the front and the guard at the back of the train to pass the signal . at what speed is the train moving at each of these times?
Answers
Answer:
Explanation:
Dear Student,
Please find below the solution to the asked query:
Given information,
Length of the train L = 100 mL = 100 m; Distance of train from signal at the time of starting d = 80 md = 80 m and acceleration of train a = 0.4 ms2/a = 0.4 ms2. Initial velocity of the train U = 0 m/sU = 0 m/s.
In order to driver to be pass the signal, train has to move "d" distance. So, time taken for driver to pass the signal from the kinematics equation is,
S = 12at2 ⇒d = 12×0.4×t2 ⇒ t = 2×800.4−−−−√ = 16004−−−−√ =400−−−√⇒t = 20 sec.S = 12at2 ⇒d = 12×0.4×t2 ⇒ t = 2×800.4 = 16004 =400⇒t = 20 sec.
Therefore, at that time train velocity is,
V = U + at ⇒ V = 0 + 0.4 × 20⇒ V = 8 m/s.V = U + at ⇒ V = 0 + 0.4 × 20⇒ V = 8 m/s.
In order to guard to pass the signal, train has to travel 180 m distance. So, the time taken by the train is,
S = 12at2 ⇒ 180 = 12×0.4×t'2 ⇒ t' = 2×1800.4−−−−−√ = 900−−−√⇒t' = 30 m/s.S = 12at2 ⇒ 180 = 12×0.4×t'2 ⇒ t' = 2×1800.4 = 900⇒t' = 30 m/s.
Therefore, the velocity of the train at this moment is,
V' = U+at = 0+0.4×30⇒ V' = 12 m/s.V' = U+at = 0+0.4×30⇒ V' = 12 m/s.
Hope this information will clear your doubts about the topic.
Regards