Question

A train starts from rest and moves with a constant acceleration on 2.0 m/s for half a minute. The brakes
are then applied and the train comes to rest in one minute after applying breaks. Find fa) the total distance
moved by the train, JK) the maximum speed attained by the train and (y the position(s) of the train
at half the maximum speed. (Assume retardation to be constant)​

Answer 1

Step-by-step explanation:

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.