QuestionAugust 6, 2025

A car accelerates from 6.00m/s to 20.0m/s This takes place over 50.0 meters. What is its rate of acceleration? 4.63m/s^2 4.36m/s^2 4.06m/s^2 3.64m/s^2

A car accelerates from 6.00m/s to 20.0m/s This takes place over 50.0 meters. What is its rate of acceleration? 4.63m/s^2 4.36m/s^2 4.06m/s^2 3.64m/s^2
A car accelerates from 6.00m/s to 20.0m/s This takes place over 50.0 meters. What is its rate of acceleration? 4.63m/s^2 4.36m/s^2 4.06m/s^2 3.64m/s^2

Solution
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Answer

3.64 \, \text{m/s}^2 Explanation 1. Identify the known values Initial velocity u = 6.00 \, \text{m/s}, final velocity v = 20.0 \, \text{m/s}, and distance s = 50.0 \, \text{m}. 2. Use the kinematic equation Apply the formula **v^2 = u^2 + 2as** to find acceleration a. 3. Solve for acceleration Rearrange to get a = \frac{v^2 - u^2}{2s}. Substitute the values: a = \frac{(20.0)^2 - (6.00)^2}{2 \times 50.0}. 4. Calculate the result a = \frac{400 - 36}{100} = \frac{364}{100} = 3.64 \, \text{m/s}^2.

Explanation

1. Identify the known values<br /> Initial velocity $u = 6.00 \, \text{m/s}$, final velocity $v = 20.0 \, \text{m/s}$, and distance $s = 50.0 \, \text{m}$.<br /><br />2. Use the kinematic equation<br /> Apply the formula **$v^2 = u^2 + 2as$** to find acceleration $a$.<br /><br />3. Solve for acceleration<br /> Rearrange to get $a = \frac{v^2 - u^2}{2s}$. Substitute the values: $a = \frac{(20.0)^2 - (6.00)^2}{2 \times 50.0}$.<br /><br />4. Calculate the result<br /> $a = \frac{400 - 36}{100} = \frac{364}{100} = 3.64 \, \text{m/s}^2$.
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