Question

A train starts from rest and moves with a constant acceleration of $2.0m/s^2$ for half a minute. the brakes are then applied and the train comes to rest in one minute. Find
the total distance moved by the train,

Answer 1

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:

$S=ut+½a{t}^2$

$=0+½\times 2\times 30\times 30$

$=900m$

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:

$v=u–at$

$=>0=60–a\times 60$

$=>a=1m/s^2$

Distance covered in this time:

$V²= u² + 2aS$

$=>0=\left(60\right)2+2(-1)S$

$=>0=3600–2S$

$=>S=3600/2=1800m$.

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:

$S=ut+½a{t}^2$

$=0+½\times 2\times 15\times 15$

$=225m$

(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:

$S=ut–½a{t}^2$

$=60x30–½x1x\left(30\right)2$

$=1800–\left(15x30\right)$

$=1800–450$

$=1350m$ (from the initial 900m covered).

So, distance from origin $=900+1350m=2250m$.