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

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

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

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

$80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec$

From first equation of motion v =u+at

So $t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec$

(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 $t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec$

(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

$a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2$

As acceleration is negative so it is a deceleration