Question

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 train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :

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

Answer 2

Explanation:

I hope you understand this solution