Solving Acceleration & Deceleration Queries for a Commuter Train Expla | StudySoup

Textbook Solutions for College Physics

Chapter 2 Problem 23

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 \(\text{m/}{{\text{s}}^{2}}\)?

Solution

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.


Title College Physics  1 
Author Paul Peter Urone, Roger Hinrichs
ISBN 9781938168000

Solving Acceleration & Deceleration Queries for a Commuter Train Expla

Chapter 2 textbook questions

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