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15). What is an object's acceleration that takes 550.0 meters to go from 6.0 m/s to 85.0 m/s?

15). What is an object's acceleration that takes 550.0 meters to go from 6.0 m/s to 85.0 m/s?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Question 15**: What is an object's acceleration that takes 550.0 meters to go from 6.0 m/s to 85.0 m/s? In this question, we need to calculate the acceleration of an object given its displacement and change in velocity. - **Initial velocity (u)** = 6.0 m/s - **Final velocity (v)** = 85.0 m/s - **Displacement (s)** = 550.0 meters We can use the kinematic equation: \[ v^2 = u^2 + 2as \] Where: - \( v \) is the final velocity, - \( u \) is the initial velocity, - \( a \) is the acceleration, - \( s \) is the displacement. Rearranging the formula to solve for \( a \) (acceleration): \[ a = \frac{v^2 - u^2}{2s} \] By substituting the given values, the acceleration can be calculated.
Transcribed Image Text:**Question 15**: What is an object's acceleration that takes 550.0 meters to go from 6.0 m/s to 85.0 m/s? In this question, we need to calculate the acceleration of an object given its displacement and change in velocity. - **Initial velocity (u)** = 6.0 m/s - **Final velocity (v)** = 85.0 m/s - **Displacement (s)** = 550.0 meters We can use the kinematic equation: \[ v^2 = u^2 + 2as \] Where: - \( v \) is the final velocity, - \( u \) is the initial velocity, - \( a \) is the acceleration, - \( s \) is the displacement. Rearranging the formula to solve for \( a \) (acceleration): \[ a = \frac{v^2 - u^2}{2s} \] By substituting the given values, the acceleration can be calculated.
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