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`        a train starts from rest and moves with a constant acceleration of 2.0 m/s^2 for half a minute. the brakes are then applied and the train comes to rest in one minute after applying breaks. find(A) the total distance moved by the train, (B) the maximum speed attained by the train, (C) the position(s)of the train at half the maximum speed. `
one year ago

## Answers : (2)

Arun
24498 Points
```							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 =0.9kmc) The position of train at half the maximum speed [30m/s]:=Let s be the position of train at half the maximum speed v²-u²=2ass=v²-u²/2as=30²-0²/2x2s=900/4=225mFor 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/s2Distance travelled during this part is :s=ut+1/2at²=60x60+1/2 (-1x60²)=3600-1800=1800m=1.8kmHence total distance travelled by train=0.9km+1.8km=2.7km
```
one year ago
Khimraj
3008 Points
```							 max speed = 2*30 = 60m/slet retardation is athen60 – a*60 = 0so a = 1m/stime when train at half the max speed30 = 2*tt = 15 secposition at time t = 15= ½*2*152 = 225 mtotal distance covered by traind = ( ½*2*302 ) + (60*60 – ½*1*602)d = (900) + (3600 – 1800) = 2700mHope it clears.
```
one year ago
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