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Grade 11General Physics

The engine of a train can give it an acceleration of 1m/s^2 and te brakes can retard it at 3m/s^2 . The minimum time in which the train can make a journey between station A ad B having a distance of 1.2 km in it. find t

Profile image of keerthi
8 Years agoGrade 11
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2 Answers

Profile image of NASRUDHEEN N
8 Years ago
The train need 1 second to accelerate 1 m/s and at the same time it only need 1 second to retard 3 m/s. So the ratio of distance travelled accelerating to distance travelled deccelerating is 3:1.
Consider distance travelling in acceleration
          displacement, s= 1.2 km ×3/4 = 900 m
        initial velocity,   u=0
           acceleration, a=1 m/s
now we want to find the time t, that is given by the equation
         s=ut+1/2 at2 =900
             0×t+1/2 ×1×t2 =900
               t2
Profile image of NASRUDHEEN N
8 Years ago
continuing answer.. sorry for inconvenience
                      t2/2=900
                          t2=450
                            t=4501/2=15× 21/2 seconds.
                      a=(v-u)/t
                        1=(v-0)/15× 21/2
gives                v=15×21/2 m/s
Now, consider distance travelling in decceleration
                         s=1.2 km×1/4=300 m
                         u=15×21/2 m/s
                         v=0
finding t by equation
         s= 1/2(u=v)t
     300=1/2(15×2½-0)t
           =(7.5×21/2)t
          t=300/(7.5×21/2)
          t=20×21/2 seconds 
total time=20×21/2+15×21/2 =35×21/2 seconds
this is the min. time

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