Question

salamat po sa makakasagot
(◠‿◠)​

Answer 1

Answer:

1)- 17.2s

Initial velocity =0m/s

Initial acceleration, a= 180×1000/3600/5= 6m/s2

 

Distance travelled in first 5 seconds:

⇒s=  1/2at2  = 1/2×6×25=75m

Velocity after 5 seconds, v=at=6×5=30m/s

Train has moved 45m during retardation and came to rest.

⇒v2 −u2 =2as

⇒0−(30×30)=(2×a×45)

⇒a=−10m/s2

 

Time taken to come to rest, t=  −10 0−30

​

=3sec

Distance travelled with uniform velocity =395−(75+45)=275m

Time taken to travel this distance =  

30

275

​

=9.2sec

Total time =9.2+3+5=17.2 s