30. A train starts from rest and moves with a constant acceleration of 2.0 m/s for hall 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) 2.7 km, 60 m/s, 225m
b) 2.2 km, 60m/s, 200m
c) 4.1 km, 50m/s, 250 m
d) None of these
Answers
EXPLANATION.
Initial velocity = u = 0
Acceleration = 2 m/s²
Final velocity = v
Time = 30 seconds.
To find
1) = Total distance moved by the train.
From Newton first equation of
kinematics.
=> v = u + at
=> v = 0 + 2 X 30
=> v = 60 m/s.
From Newton second equation of
kinematics.
=> s1 = ut + 1/2 at²
=> s1 = 0 + 1/2 X 2 X 30 X 30
=> s1 = 900 m
when break are applied.
=> u = 60 m/s
=> v = 0
=> Time = 60 seconds = 1 minute.
Deceleration = a = ( v - u) / t
=> a = 0 - 60 / 60
=> a = -1 m/s²
From Newton third equation of
kinematics.
=> v² = u² + 2as2
=> (0)² = (60)² + 2 X (-1) s2
=> s2 = 1800 m
=> Total distance = S1 + s2
=> 900 + 1800 = 2700 m = 2.7 km
To find.
2) = The maximum speed attend by the
train.
=> v = 60 m/s
To find
3) = The train at half the maximum
speed.
Half the maximum speed = 60/2 = 30 m/s
initial velocity = 0
From Newton third equation of
kinematics.
=> v² = u² + 2as
=> (30) ² = 0 + 2 X 2 X S
=> s = 225 m From the starting point.
Therefore,
=> u = 60 m/s
=> v = 30 m/s
=> a = -1 m/s²
Therefore,
=> v² = u² + 2as
=> (30) ² = (60) ² + 2 X (-1) X S
=> 900 = 3600 - 2s
=> s = 1350 m
Position are = 900 + 1350 = 2250 m
=> 2.25 km from starting point.