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A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

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6 years ago

```							Sol. Initial velocity u = 0
Acceleration a = 2 m/s2. Let final velocity be v (before applying breaks)
t = 30 sec
v = u + at ⇒ 0 + 2 × 30 = 60 m/s
a) S1 = ut + ½ at2 = 900 m
when breaks are applied u’ = 60 m/s
v’ = 0, = 60 sec (1 min)
Declaration a’ = (v-u)/t = = (0 – 60 ) / 60 = - 1 m/s2.
S2 = (V^'2-u^(;2))/(2a^' ) = 1800 m
Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km.
b) The maximum speed attained by train v = 60 m/s
c) Half the maximum speed = 60/2= 30 m/s
Distance S =  (V^2-u^2)/2a = (〖30〗^2-0^2)/(2 x 2) = 225 m from starting point
When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s.
∴ u = 60 m/s, v = 30 m/s, a = –1 m/s2
Distance = (V^2-u^2)/2a = (〖30〗^2-〖60〗^2)/(2(-1)) = 1350 m
Position is 900 + 1350 = 2250 = 2.25 km from starting point

```
6 years ago
```							Sol. Initial velocity u = 0 Acceleration a = 2 m/s2. Let final velocity be v (before applying breaks) t = 30 sec v = u + at ⇒ 0 + 2 × 30 = 60 m/s a) S1 = ut + ½ at2 = 900 m when breaks are applied u’ = 60 m/s v’ = 0, = 60 sec (1 min) Declaration a’ = (v-u)/t = = (0 – 60 ) / 60 = - 1 m/s2. S2 = (V^'2-u^(;2))/(2a^' ) = 1800 m Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s c) Half the maximum speed = 60/2= 30 m/s Distance S = (V^2-u^2)/2a = (〖30〗^2-0^2)/(2 x 2) = 225 m from starting point When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s. ∴ u = 60 m/s, v = 30 m/s, a = –1 m/s2 Distance = (V^2-u^2)/2a = (〖30〗^2-〖60〗^2)/(2(-1)) = 1350 m Position is 900 + 1350 = 2250 = 2.25 km from starting point
```
one year ago
```							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 speedv²-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 month ago
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