Question

A train (P) acceleration from rest A at a ms
","s, maintains a steady speed for T s, and
comes to rest at B by retarding uniformly for
further 21 s. Another train (Q) starts at "A"
and accelerates uniformly to reach B in the
same time. The minimum time, taken by a
third train that starts at A and stops at B is.
It can have a top speed of "at "ms!
За
maximum acceleration of 2 and
maximum retardation of a ms?
nis
(71 +37)
3
(1 1137)
3
(51 +37)
3
(91 +3T)
3​

Answer 1

Given,

Velocity v

1

achieve in time t

1

,due to acceleration, a

1

=0.2ms

−2

v

1

=u

1

+at

1

⇒v

1

=0+0.2t

1

Velocity reduce to zero in time t

2

, due to retardation, a

2

=−0.4ms

−2

v

2

=v

1

+at

2

⇒0=v

1

−0.4t

2

Add time

t

1

+t

2

=

0.2

v

1

+

0.4

v

1

⇒30×60=v

1

×7.5

⇒v

1

=240m/s

Apply kinematic equation

v

2

−u

2

=2as

s=

2a

v

2

−u

2

Total distance S

1

+S

2

=

2a

1

v

1

2

+

2a

2

v

1

2

=

2×0.2

240

2

+

2×0.4

240

2

=216km