QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
During a 5 -second time interval, the average acceleration $a$, in meters per second squared, of an object with an initial velocity of 12 meters per second is defined by the equation $a=\frac{v f- 12}{5}$, where $v f$ is the final velocity of the object in meters per second. If the equation is rewritten in the form $v f=x a+y$, where $x$ and $y$ are constants, what is the value of $x$ ?
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Answer
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Step 1:: Recall that the equation given is $a = \frac{v_f - 12}{5}$.
To rewrite this equation in the form $v_f = xa + y$, we need to isolate $v_f$.
Step 2:: To isolate $v_f$, first multiply both sides of the equation by 5.
This gives us $5a = v_f - 12$.
Step 3:: Now, add 12 to both sides of the equation to get $5a + 12 = v_f$.
Step 4:: Comparing this equation to the desired form $v_f = xa + y$, we can identify that $x = 1$ and $y = 1$.
Final Answer
The value of $x$ is 5.
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