A train starts from rest and accelerates uniformlyat a rate of 2 m s2

1.	A train starts from rest and accelerates uniformlyat a rate of 2 m s2 for 10 s. It then maintains aconstant speed for 200 s. The brakes are thenapplied and the train is uniformly retarded andcomes to rest in 50 s. Find: (i) the maximumvelocity reached, (ii) the retardation in the last50 s, (iii) the total distance travelled, and(iv) the average velocity of the train.15.
Answer» In this Journey, the train has accelerated and decelerated motion. For accelerated motion :Initial velocity=u=0m/sacceleration=a=2m/s2time=t=half minute=30 secFinal velocity =V=?From first equation of motion , v=u+atv=0+2x30=60m/sSo train has travelled 60m/s before brakes are appplied. Hence the maximum speed attained by train 60m/s. Distance travelled : From second equation of motion : S=ut+1/2at²S=0x30+1/2 x 2x 30²s=900m/ss=900/1000 =0.9km c) The position of train at half the maximum speed [30m/s]:=Let s be the position of train at half the maximum speedv²-u²=2ass=v²-u²/2as=30²-0²/2x2s=900/4=225m For decelerated motion :Final Velocity=V=o m/sinitial velocity=u=60m/stime=60secFrom first equation of motion : v=u+at0=60+ax60a=-60/60=-1m/s2 Distance travelled during this part is :s=ut+1/2at²=60x60+1/2 (-1x60²)=3600-1800=1800m=1.8km Hence total distance travelled by train=0.9km+1.8km=2.7km -

A train starts from rest and accelerates uniformlyat a rate of 2 m s2 for 10 s. It then maintains aconstant speed for 200 s. The brakes are thenapplied and the train is uniformly retarded andcomes to rest in 50 s. Find: (i) the maximumvelocity reached, (ii) the retardation in the last50 s, (iii) the total distance travelled, and(iv) the average velocity of the train.15.

Answer»

In this Journey, the train has accelerated and decelerated motion.

For accelerated motion :Initial velocity=u=0m/sacceleration=a=2m/s2time=t=half minute=30 secFinal velocity =V=?From first equation of motion , v=u+atv=0+2x30=60m/sSo train has travelled 60m/s before brakes are appplied.

Hence the maximum speed attained by train 60m/s.

Distance travelled : From second equation of motion : S=ut+1/2at²S=0x30+1/2 x 2x 30²s=900m/ss=900/1000 =0.9km

c) The position of train at half the maximum speed [30m/s]:=Let s be the position of train at half the maximum speedv²-u²=2ass=v²-u²/2as=30²-0²/2x2s=900/4=225m

For decelerated motion :Final Velocity=V=o m/sinitial velocity=u=60m/stime=60secFrom first equation of motion : v=u+at0=60+ax60a=-60/60=-1m/s2

Distance travelled during this part is :s=ut+1/2at²=60x60+1/2 (-1x60²)=3600-1800=1800m=1.8km

Hence total distance travelled by train=0.9km+1.8km=2.7km

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