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Grade: 11
        
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 month ago

Answers : (2)

Arun
20803 Points
							
In this Journey, the train has accelerated and decelerated motion.
For accelerated motion :
Initial velocity=u=0m/s
acceleration=a=2m/s2
time=t=half minute=30 sec
Final velocity =V=?
From first equation of motion , v=u+at
v=0+2x30=60m/s
So 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+1/2at²
S=0x30+1/2 x 2x 30²
s=900m/s
s=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 speed 
v²-u²=2as
s=v²-u²/2a
s=30²-0²/2x2
s=900/4=225m
For decelerated motion :
Final Velocity=V=o m/s
initial velocity=u=60m/s
time=60sec
From first equation of motion : v=u+at
0=60+ax60
a=-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
 
one month ago
Khimraj
2788 Points
							
 
max speed = 2*30 = 60m/s
let retardation is a
then
60 – a*60 = 0
so a = 1m/s
time when train at half the max speed
30 = 2*t
t = 15 sec
position at time t = 15
= ½*2*152 = 225 m
total distance covered by train
d = ( ½*2*302 ) + (60*60 – ½*1*602)
d = (900) + (3600 – 1800) = 2700m
Hope it clears.
28 days ago
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