Question

(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 ?

Answer 1

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration =
`2.677m/sec^2`

Step-by-step explanation:

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

Acceleration
`a=1.35m/sec^2`

Final speed v = 80 km/hr

`80km/hr=(80* 1000)/(3600sec)=22.22m/sec`

From first equation of motion v =u+at

So
`t=(v-u)/(a)=(22.22-0)/(1.35)=16.46 sec`

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

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

Deceleration
`a=1.65m/sec^2`

So
`t=(v-u)/(a)=(0-22.22)/(-1.65)=13.466 sec`

(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

`a=(v-u)/(t)=(0-22.22)/(8.3)=-2.6771m/sec^2`

As acceleration is negative so it is a deceleration