Answer:
(a) Time t = 16.46 sec
(b) Time t =13.466 sec
(c) Deceleration = [tex]2.677m/sec^2[/tex]
Explanation:
(a) As the train starts from rest its initial velocity u = 0 m/sec
Acceleration [tex]a=1.35m/sec^2[/tex]
Final speed v = 80 km/hr
[tex]80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec[/tex]
From first equation of motion v =u+at
So [tex]t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec[/tex]
(b) Now initial speed u = 22.22 m/sec
As finally train comes to rest so final speed v=0 m/sec
Deceleration [tex]a=1.65m/sec^2[/tex]
So [tex]t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec[/tex]
(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec
Final velocity v = 0 m/sec
Time t = 8.30 sec
So acceleration is given by
[tex]a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2[/tex]
As acceleration is negative so it is a deceleration
