Question

The Speed Of A Train Is Reduced From 80km/h To 40km/h In A Distance Of 500m On Applying The Brakes (I)how Much Further Will The Train Travel Before Coming To Rest (II)assuming The Retardation Remain Constant,how Long Will It Take To Bring The Train To Rest After The Application Of The Brakes

Answer 1

average speed during initial slowdown = (80+40)/2 = 60km/h
60 km/h * t = 0.5 km
t = 5/600 hours = (1/120)hr
(1 hr/120)(60min/hr)(60s/min)
= 30 seconds
so
a = change in speed/time
= -40 km/hr /30 s
= (-40 * 1000 m/3600s)/30s
= -(4/3)/3.6 m/s^2
so
Vi = (40/3.6) m/s
final v = 0 =Vi+at
time to stop = -Vi/a
= (40/3.6) / [(4/3)/3.6]
=10*3 seconds
=30 seconds more
distance in final phase =(20/3.6)m/s(30s) =600/3.6 = 167 m more
total time = 30+30 = 60 seconds

Answer 2

I don't get the calculations

Answer 3

I need the answer to this question

Answer 4

Is not correct

Answer 5

I don't get the solution

Answer 6

I am confused

Answer 7

Explain please

Answer 8

It not clear🤷

Answer 9

This is not explainatry