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Solving Acceleration & Deceleration Queries for a Commuter Train Expla

Chapter 2, Problem 23

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QUESTION:

(a) A light-rail commuter train accelerates at a rate of \(1.35\ \text{m/}{{\text{s}}^{2}}\). How long does it take to reach its top speed of \(80.0\ \text{km/h}\), starting from rest?

(b) The same train ordinarily decelerates at a rate of \(1.65\ \text{m/}{{\text{s}}^{2}}\). 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 \(80.0\ \text{km/h}\) in 8.30 s. What is its emergency deceleration in \(\text{m/}{{\text{s}}^{2}}\)?

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QUESTION:

(a) A light-rail commuter train accelerates at a rate of \(1.35\ \text{m/}{{\text{s}}^{2}}\). How long does it take to reach its top speed of \(80.0\ \text{km/h}\), starting from rest?

(b) The same train ordinarily decelerates at a rate of \(1.65\ \text{m/}{{\text{s}}^{2}}\). 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 \(80.0\ \text{km/h}\) in 8.30 s. What is its emergency deceleration in \(\text{m/}{{\text{s}}^{2}}\)?

ANSWER:

Step 1 of 4

Consider the given data as follows.

The acceleration of a train is \(a=1.35\ \text{m/}{{\text{s}}^{2}}\).

The final speed of the train is \(v=80.0\ \text{km/h}\times \dfrac{{{10}^{3}}\ \text{m}}{3600\ \text{s}}=22.2\ \text{m/s}\).

The deacceleration of the train is \(a=1.65\ \text{m/}{{\text{s}}^{2}}\).

 

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Solving Acceleration & Deceleration Queries for a Commuter Train Expla
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Explore the physics of a light-rail commuter train's motion. Understand how to calculate times for acceleration and deceleration using the equations of motion. Dive into the concept of emergency deceleration and its implications.


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