How does that equal to 56.25 m? The image contains a physics problem involving the kinematic equations of motion:1. **Equation**: \[ V_f^2 = V_i^2 + 2a \Delta x \] - \( V_f \) is the final velocity. - \( V_i \) is the initial velocity. - \( a \) is the acceleration. - \( \Delta x \) is the displacement.2. **Rearranging to Solve for Displacement**: \[ \Delta x = \frac{V_f^2 - V_i^2}{2a} \]3. **Substituting Values**: \[ \Delta x = \frac{(0 \, \text{m/s})^2 - (30.0 \, \text{m/s})^2}{2(-8.00 \, \text{m/s}^2)} \]4. **Calculation**: \[ \Delta x = 56.25 \, m \]This calculation shows the displacement (\(\Delta x\)) of an object decelerating from an initial velocity of 30.0 m/s to a final velocity of 0 m/s with an acceleration of -8.00 m/s². The result is a displacement of 56.25 meters.

Answered: V? = V? + 2a A æ Ax= Ax=V° - V;? | 2a…

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V? = V? + 2a A æ Ax= Ax=V° - V;? | 2a (0 m/s)² –(30.0 m/s)² 2(-8,00 m/s²) = 56. 25 m

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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