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CBSE Class 11-science Answered

Please answer question number 27 .  
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Asked by arunavamitra50 | 19 Jun, 2018, 01:31: PM
answered-by-expert Expert Answer
Let the train travelled a distance S1 in time t1 seconds with acceleration 0.2 m/s2   .
 
S1 = (1/2)×0.2×t12 ....................(1)
 
Let train speed is u after t1 seconds. u = 0.2×t1 ..............(2)
 
Let the train travelled a distance S2 in time t2 seconds with retardation 0.4 m/s2   .
 
S2 = u×t2 - (1/2)×0.4×t22  = 0.2×t1×t2 - (1/2)×0.4×t22    ...........................(3)
 
we have the equation of motion " v = u - a×t ", where v is final speed and u is initial speed, a is retardation and t is time.
In this equation if the moving object comes to rest after t seconds, then we put v = 0 to get the time from initial speed and retardation.
 
hence t2 = (0.2×t1) / 0.4 = t1 / 2......................(4)
 
we are given that  t1 + t2 = 60×30 = 1800 .............(5)
 
solving eqns.(4) and (5), we get t1 = 1200 s and t2 = 600 s
 
we can calculate S1 and S2 from eqns.(1) and (3) using the respective values of t1 and t2.
 
we get S1 = 144 km and S2 = 72 km . Hence total distance S1+S2 = 216 km
Answered by Thiyagarajan K | 19 Jun, 2018, 03:20: PM
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