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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.

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.

Grade:upto college level

4 Answers

Deepak Patra
askIITians Faculty 471 Points
9 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
Shri sinha
16 Points
4 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
ankit singh
askIITians Faculty 614 Points
3 years ago
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
Patel
13 Points
3 years ago
Train starts from rest, hence the initial velocity u = 0.
 
It moves with acceleration = 2m/s2 for half minute (30 seconds).
 
Distance covered in this time interval is given by:
 
S=ut+½at 
2
 
 
=0+½×2×30×30
 
=900m
 
Velocity attained by this acceleration after 30 seconds:
 
v=u+at
 
=>v=0+2x30
 
=>v=60m/s
 
From this velocity, brakes are applied and train comes to rest in 60 seconds.
 
The retardation is given by:
 
v=u–at
 
=>0=60–a×60
 
=>a=1m/s 
2
 
 
Distance covered in this time:
 
$$V2= u2 + 2aS$$
 
=>0=(60)2+2(−1)S
 
=>0=3600–2S
 
=>S=3600/2=1800m.
 
So, total distance moved =900m+1800m=2700m.
 
Maximum speed of the train=60m/s.
 
Position of the train at half its maximum speed.
 
Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s.
 
(I) When the train is accelerating with an acceleration of 2 m/s,
 
time at which speed = 30m/s is:
 
v=u+at
 
=>30=0+2xt
 
=>t=15s
 
At 15s, distance covered from origin is:
 
S=ut+½at 
2
 
 
=0+½×2×15×15
 
=225m
 
(II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is:
 
v=u–at
 
=>30=60–1xt
 
=>t=30s
 
At 30s, distance covered is:
 
S=ut–½at 
2
 
 
=60x30–½x1x(30)2
 
=1800–(15x30)
 
=1800–450
 
=1350m (from the initial 900m covered).
 
So, distance from origin =900+1350m=2250m.

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