The acceleration of a vehicle can berepresented by the followingequation : ( dVldt = 2 - 0.05 V), where(V) is the vehicle speed in (m/sec). Ifthe vehicle is traveling at (72 km/hr),determine the distance after (4seconds) of acceleration. *O Distance = 158 mO Distance =169 mO Distance =183 mO Distance = 166 mO Distance = 160 m%3D

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Description: The acceleration of a vehicle can berepresented by the followingequation : ( dVldt = 2 - 0.05 V), where(V) is the vehicle speed in (m/sec). Ifthe vehicle is traveling at (72 km/hr),determine the distance after (4seconds) of acceleration. *O Distance

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The acceleration of a vehicle can be represented by the following equation : ( dVldt = 2 - 0.05 V ), where (V) is the vehicle speed in (m/sec). If the vehicle is traveling at (72 km/hr), %3D determine the distance after (4 seconds) of acceleration. *

The acceleration of a vehicle can be represented by the following equation : ( dVldt = 2 - 0.05 V ), where (V) is the vehicle speed in (m/sec). If the vehicle is traveling at (72 km/hr), %3D determine the distance after (4 seconds) of acceleration. *

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 55P: Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm left ventricle of the heart....

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