A train starts from rest and moves with a constant accelerationof 2.0`m/s^2` 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.

Explanation

Initial velocity
u=0
Acceleration
`a=2m/s^2
Let finl velocity be v
(before applying breaks)
`t=30 sec
v=u+at
=0+2xx30
= 60 m/s
a. S_1 = ut+1/2 at^2
= 1/2xx2xx(30)^2
=900m
When breaks are applied
u'=60 m/s
v'=0
t=60 sec (1 min) Deceleration
a'+(v-u)/t
=(0-60)/60=-1m/s^2
S_2= (v^2-u^2)/(2a')
=(0^2-60^2)/(2(-1))ltbrge 1800m
total S=S_1+S_2
=1800+900
=2700m
=2.7m`
b. The maximum speed asttained by train
v=60m/s
c. Half rthe maximum speed
`=60/2=30 m/s
Distance
S (v^2-u^2)/(2a')ltbrge (30^2-60^2_)/(2(-1))ltbrge1350 m
Position= 900+1350, 2250
=2.25 km` from starting point