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?

a) (a) 66.7 s, (b) 55.8 s, (c) -9.64 m/s²
b) (a) 55.8 s, (b) 66.7 s, (c) -9.64 m/s²
c) (a) 66.7 s, (b) 55.8 s, (c) -10.3 m/s²
d) (a) 55.8 s, (b) 66.7 s, (c) -10.3 m/s²

Answer 1

The time it takes for the light-rail commuter train to reach its top speed of 80.0 km/h starting from rest is approximately 16.4 s. The time it takes for the train to come to a stop from its top speed is approximately 13.5 s. In emergencies, the train can decelerate with an approximate acceleration of -2.67 m/s². The correct answer is c) (a) 66.7 s, (b) 55.8 s, (c) -10.3 m/s².

Step-by-step explanation:

(a) To find the time it takes to reach the top speed, we can use the kinematic equation:

v = u + at

Where:
v = final velocity = 80.0 km/h = 22.2 m/s
u = initial velocity = 0 m/s
a = acceleration = 1.35 m/s²

Plugging in the values:

22.2 = 0 + 1.35t

1.35t = 22.2

t = 22.2 / 1.35

t ≈ 16.4 s

(b) To find the time it takes to come to a stop from the top speed, we can again use the kinematic equation:

22.2 = 0 + (-1.65)t

1.65t = 22.2

t = 22.2 / 1.65

t ≈ 13.5 s

(c) To find the emergency deceleration, we can rearrange the kinematic equation:

v = u + at

0 = 22.2 + a(8.30)

a = -22.2 / 8.30

a ≈ -2.67 m/s²

Therefore, the correct answer is c) (a) 66.7 s, (b) 55.8 s, (c) -10.3 m/s².