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Grade: upto college level

                        

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.

6 years ago

Answers : (3)

Deepak Patra
askIITians Faculty
471 Points
							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
Shri sinha
16 Points
							
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
ankit singh
498 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
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