Question

plzzz answer me fast.......

Answer 1

hey buddy
Suppose the distance travelled by both the trains before they stop =s
For train 1,
u=72 km/h=72×1000/60×60m/s=20 m/s
a=−1ms^2
Using relation,
v^2−u^2=2as
0−20^2=2×(−1)×s
400=2s
s=200 m
For train 2,
u=90 km/h=90×1000/60×60=25 m/s
a=−1m/s^2
v^2−u^2=2as
0−25^2=2×(−1)×s
s=312.5 m
Since they are 1 km i.e. 1000 m apart and sum of both their distance is less than 1000 m, they would stop before time.
There would no collision
hope it will help

Answer 2

Hey friend.... .

let the first tain be A

&. 2nd train be B...

For train A

u1= 72kmph=20m/s
a = -1 m/s²
& v1=0 [As train stops lastly]

s 1= (v²-u²)/2a
= 0-20²/2*(-1)
=200meter..

now for train B

u2= 90kmph= 25 m/s

v2= 0 (as train stops lastly)

a=-1m/s²

s 2= v²-u²/2a

=0-25²/2*(-1)

= 2
625/2

=312.50 meter

now
s1+s2
= 200+ 312.50 meter

512.50 meter

And given distance between them=1000m(1km)
when break applied. which is greater than travelled distance by train A&B

hence trains do not collide...

Hope this will help you. .