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 accelerates opposite to the motion 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 accelerate opposite to the motion more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency acceleration in meters per second squared?

a) (a) 99 s, (b) 48.48 s, (c) 2.25 m/s²
b) (a) 64 s, (b) 60.61 s, (c) 2.50 m/s²
c) (a) 59 s, (b) 54.55 s, (c) 3.00 m/s²
d) (a) 69 s, (b) 72.73 s, (c) 3.75 m/s²

Answer 1

The train takes 16.46 seconds to reach its top speed of 80.0 km/h from rest, takes 13.47 seconds to come to a stop from its top speed ordinarily, and in emergencies, it decelerates at a rate of 2.68 m/s² to come to rest from 80.0 km/h in 8.30 seconds.

Step-by-step explanation:

The question pertains to the kinematics of a light-rail commuter train and involves calculating time taken to reach top speed, time taken to decelerate, and emergency deceleration rate.

1. (a) To find out how long it takes for the train to reach its top speed of 80.0 km/h (which is 22.22 m/s), starting from rest, we use the formula t = v/a where v is the final velocity and a is the acceleration. This gives us t = 22.22 m/s / 1.35 m/s² = 16.46 s.
2. (b) To find the time taken to come to a stop from top speed, we again use the formula t = v/a, with a being the deceleration rate. So, t = 22.22 m/s / 1.65 m/s² = 13.47 s.
3. (c) For the emergency deceleration, we know the time and initial velocity, and we need to find the acceleration. Using a = v/t, we get a = 22.22 m/s / 8.30 s = 2.68 m/s², which is the emergency deceleration rate.

By applying these kinematic equations, we calculate the various aspects of the train's motion under different circumstances.