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Shane Macguire 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.
3 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
474 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
3 years ago
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