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1. Nurmber \& Quantity Plus Modeling * A car accelerated from 88 feet per second (fps) to 220 fps in exactly 3 seconds. Assuming the acceleration was constant, what was the car's acceleration, in feet per second per second, from 88 fps to 220 fps? F. \( \quad \frac{1}{44} \) G. \( 29 \frac{1}{3} \) H. 44 J. \( 75 \frac{1}{3} \) K. \( 102 \frac{2}{3} \)

Ask by Malone Schwartz. in the United States
Feb 03,2025

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The car's acceleration is 44 feet per second squared. **Answer:** H. 44

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To determine the car's constant acceleration, we can use the formula for acceleration: \[ a = \frac{\Delta v}{\Delta t} \] Where: - \(\Delta v\) is the change in velocity. - \(\Delta t\) is the change in time. Given: - Initial velocity (\(v_0\)) = 88 feet per second (fps) - Final velocity (\(v\)) = 220 fps - Time interval (\(\Delta t\)) = 3 seconds Calculate the change in velocity: \[ \Delta v = v - v_0 = 220 \, \text{fps} - 88 \, \text{fps} = 132 \, \text{fps} \] Now, calculate the acceleration: \[ a = \frac{132 \, \text{fps}}{3 \, \text{s}} = 44 \, \text{feet per second squared (fps²)} \] **Answer:** **H. \(44\)**

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To find the car's acceleration, we can use the formula for acceleration, which is the change in velocity divided by the time taken. Here, the initial velocity (u) is 88 fps and the final velocity (v) is 220 fps, with a time (t) of 3 seconds. Acceleration (a) is thus calculated as: \[ a = \frac{v - u}{t} = \frac{220 \, \text{fps} - 88 \, \text{fps}}{3 \, \text{s}} = \frac{132 \, \text{fps}}{3 \, \text{s}} = 44 \, \text{fps}^2. \] So, the answer is **H. 44**. Acceleration in physics isn’t just a figure; it’s also an exciting moment in any race! Remember, constant acceleration means the speed increases uniformly over time, which is exactly how race cars maintain their fierce competition. Next time you see cars zooming around a track, think about how their speed changes from just a simple equation! A little bit of math can explain those thrilling moments of acceleration that make racing so exhilarating.

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