Hall Adkins
03/03/2023 · Middle School
\[ v^{2}=u^{2}+2 a s \] Where \( v \) is the final velocity (in \( \mathrm{m} / \mathrm{s} \) ), \( u \) is the initial velocity (in \( \mathrm{m} / \mathrm{s} \) ), \( a \) is the acceleration (in \( \mathrm{m} / \mathrm{s}^{2} \) ) and \( s \) is the distance (in meters). Find \( v \) when \( u \) is \( 9 \mathrm{~m} / \mathrm{s}, a \) is \( 7 \mathrm{~m} / \mathrm{s}^{2} \), and \( s \) is 28 meters. A. \( \sqrt{178} \mathrm{~m} / \mathrm{s} \) B. \( 20 \mathrm{~m} / \mathrm{s} \) C. \( \sqrt{473} \mathrm{~m} / \mathrm{s} \) D. \( \sqrt{410} \mathrm{~m} / \mathrm{s} \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To find the final velocity \( v \) using the equation
\[
v^{2} = u^{2} + 2 a s,
\]
we can substitute the given values:
- \( u = 9 \, \mathrm{m/s} \)
- \( a = 7 \, \mathrm{m/s}^2 \)
- \( s = 28 \, \mathrm{m} \)
First, we calculate \( u^{2} \):
\[
u^{2} = (9 \, \mathrm{m/s})^{2} = 81 \, \mathrm{m}^2/\mathrm{s}^2.
\]
Next, we calculate \( 2 a s \):
\[
2 a s = 2 \times 7 \, \mathrm{m/s}^2 \times 28 \, \mathrm{m} = 14 \times 28 = 392 \, \mathrm{m}^2/\mathrm{s}^2.
\]
Now, we can substitute these values into the equation:
\[
v^{2} = 81 \, \mathrm{m}^2/\mathrm{s}^2 + 392 \, \mathrm{m}^2/\mathrm{s}^2 = 473 \, \mathrm{m}^2/\mathrm{s}^2.
\]
Finally, we take the square root to find \( v \):
\[
v = \sqrt{473} \, \mathrm{m/s}.
\]
Thus, the answer is:
C. \( \sqrt{473} \, \mathrm{m/s} \)
Quick Answer
The final velocity \( v \) is \( \sqrt{473} \, \mathrm{m/s} \). Therefore, the correct answer is C. \( \sqrt{473} \, \mathrm{m/s} \).
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