An object moving vertically is at the given heights at the specified times. Find the position equation $s = \frac { 1 } { 2 } a t ^ { 2 } + v _ { o } t + s _ { o }$ for the object.
At $t = 1$ second, $s = 234$ feet
At $t = 2$ seconds, $s = 202$ feet
At $t = 3$ seconds, $s = 138$ feet
A) $s = - 32 t ^ { 2 } + 16 t + 234$
B) $s = - 16 t ^ { 2 } + 16 t - 234$
C) $s = - 16 t ^ { 2 } + 16 t + 234$
D) $s = - 8 t ^ { 2 } + t - 234$
E) $s = - 16 t ^ { 2 } - 16 t + 234$

Question

Quizplus · Accepted Answer

Using the given information, we can plug in the values and get the following system of equations:
\begin{align*}
234&=\frac{1}{2}a(1)^2+v_o(1)+s_o\
202&=\frac{1}{2}a(2)^2+v_o(2)+s_o\
138&=\frac{1}{2}a(3)^2+v_o(3)+s_o
\end{align*}

Simplifying each equation, we get:
\begin{align*}
234&=\frac{1}{2}a+v_o+s_o\
202&=2a+2v_o+s_o\
138&=\frac{9}{2}a+3v_o+s_o
\end{align*}

Now, we have three equations with three unknowns. Solving for $a$, $v_o$, and $s_o$, we get:

$a=-16$, $v_o=16$, $s_o=234$

Substituting these values into the position equation, we get:

$s=-16t^2+16t+234$

Therefore, the answer is choice $	extbf{(C)}$.

An Object Moving Vertically Is at the Given Heights at the Specified