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The equation of position vs. time for a moving object, in SI units, is as \(x=-t^{2}+6 t-9 .\) Which of the following choices are correct?

(a) The object's acceleration is constant and its magnitude is \(1 \mathrm{~m} / \mathrm{s}^{2}\).
(b) The object's velocity at the initial time \(t=0\) is to the negative \(x\)-axis.
(c) The object's initial position is on the negative side of the \(x\)-axis.
in Physics & Chemistry by Platinum (130,538 points) | 90 views

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(a) This equation has a quadratic form so its acceleration is constant. Comparing this equation with standard constant acceleration kinematic equation, \(x=\frac{1}{2} a t^{2}+v_{0} t+x_{0}\), we will find its magnitude as
\[
\frac{1}{2} a=-1 \Rightarrow a=-2 \mathrm{~m} / \mathrm{s}^{2}
\]
So this choice is incorrect.

(b) "Object's velocity at the initial time" means its initial velocity. Comparing the two equations below reveals that initial velocity is \(v_{0}=+6 \mathrm{~m} / \mathrm{s}\).
\[
\begin{gathered}
x=\frac{1}{2} a t^{2}+v_{0} t+x_{0} \\
x=-t^{2}+6 t-9
\end{gathered}
\]
The positive sign indicates that the initial velocity is toward the positive \(x\)-axis. So, this choice is also incorrect.

(c) By setting \(t=0\) in the position-time equation, its initial position is obtained. So, \(x_{0}=\) \(-9 \mathrm{~m}\). Negative indicates that the object is on the negative side of the \(x\)-axis initially. Thus, this choice is correct.
by Platinum (130,538 points)

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