The acceleration of a vehicle can be represented by the following equation : ( dV/dt = 2 - 0.05 V), where (V) is the vehicle speed in (m/sec). If the vehicle is traveling at (72 km/hr), determine the distance after (4 seconds) of acceleration. * Distance =183 m Distance = 160 m %3D Distance = 158 m Distance =169 m Distance = 166 m %3D

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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The acceleration of a vehicle can
be represented by the following
equation : (dV/dt = 2 - 0.05 V),
where (V) is the vehicle speed in
(m/sec). If the vehicle is traveling
at (72 km/hr), determine the
distance after (4 seconds) of
acceleration.
Distance =183 m
O Distance = 160 m
Distance
= 158 m
Distance =169 m
Distance = 166 m
Transcribed Image Text:The acceleration of a vehicle can be represented by the following equation : (dV/dt = 2 - 0.05 V), where (V) is the vehicle speed in (m/sec). If the vehicle is traveling at (72 km/hr), determine the distance after (4 seconds) of acceleration. Distance =183 m O Distance = 160 m Distance = 158 m Distance =169 m Distance = 166 m
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