Question

answer fast it's important ​

Answer 1

Answer:

Using equation of motion for calculating the time in which the train is brought to rest:

v 2=u 2 +2as

⇒a=−54000km/h

20 2

=90

2 +2a(75×10 −3 )v=u+at

0=90+(−54000)t

⇒t=0.00166h=0.00166×60min=0.1min=0.1×60s=6s

So, the train comes to rest in 6 s.

Distance traveled in the first half, i.e., distance traveled in 3s:

v=90+(−54000)( 36003 )=45km/h

(45)

2 =90

2 +2(−54000)s

s=56.25m

Thus, distance travelled in second half = (75-56.25)m = 18.75 m.

hope it helps