(a) a light-rail commuter train accelerates at a rate of 1.35 m/s2 . how long does it take to reach its top speed of 80.0 km/h, starting from rest?
(b) the same train ordinarily decelerates at a rate of 1.65 m/s2 . how long does it take to come to a stop from its top speed?
(c) in emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. what is its emergency deceleration in m/s2 ?

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration =

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration

Final speed v = 80 km/hr

From first equation of motion v =u+at

So

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration

So

(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

As acceleration is negative so it is a deceleration

higher potential

explanation:

because it is in a high speed so it must move into a of high potential