Question

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? 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? 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:

16.46 seconds.

13.46 seconds

2.67 m/s²

Explanation:

Acceleration = a = 1.35 m/s²

Final velocity = v = 80 km/h = $80\frac{1000}{3600}=\frac{200}{9}\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow \frac{200}{9}=0+1.35t\\\Rightarrow t=\frac{\frac{200}{9}}{1.35}=16.46\ s$

Time taken to accelerate to top speed is 16.46 seconds.

Acceleration = a = -1.65 m/s²

Initial velocity = u = 80 km/h= $80\frac{1000}{3600}=\frac{200}{9}\ m/s$

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=\frac{200}{9}-1.65t\\\Rightarrow t=\frac{\frac{200}{9}}{1.65}=13.46\ s$

Time taken to stop the train from top speed is 13.46 seconds

Initial velocity = u = 80 km/h= $80\frac{1000}{3600}=\frac{200}{9}\ m/s$

Time taken = t = 8.3 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=\frac{200}{9}+a8.3\\\Rightarrow a=\frac{-\frac{200}{9}}{8.3}=-2.67\ m/s^2$

Emergency deceleration is 2.67 m/s²