Question: A train slows down as it rounds a sharp horizontal turn, going from 90.0km/h to50.0km/h in the 15.0s it takes to round the bend. The radius of the curve is 150m. Compute thetotal acceleration at the moment the train speed reaches 50.0km/h. Assume the train continuesto slow down at this time at the same rate.A. a-2 m/s?B. a-0.74 m/s?C. a=1.48 m/s?D. a=1.29 m/s?E. a=2 m/s?

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University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter3: Motion Along A Straight Line

Section: Chapter Questions

Problem 72P: (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is...

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Concept explainers

Special Relativity

The theory of special relativity discusses how space and time are related to objects traveling along a straight line at a constant speed. Simply stated, as an object reaches the speed of light, its mass becomes infinite and it is unable to move faster than light. This interstellar speed limit has been the topic of much debate in astronomy, and also in science fiction when people consider how to fly over large distances.

Question

Expert Solution

Step 1

Given:

A train goes from $(v_{1})$ $90 k m / h t o (v_{2}) 50 k m / h$

The Duration of the train travel = 15 s

The radius of the curve =50 km/h

Step 2

On calculating the tangential acceleration using the formula as follows,

$a = \frac{v_{2} - v_{1}}{t}$

On substituting the given values in the tangential formula,

$a = \frac{\frac{50 \times 1000}{3600} - \frac{90 \times 1000}{3600}}{15}$

$a_{\tan} = - 0.7407 m / s^{2}$

The Tangential acceleration is $- 0.7407 m / s^{2}$

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