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

The object is moving vertically, so the acceleration due to gravity, $g = -32$ ft/s^2, is the only acceleration acting on the object. The position equation is $s = \frac { 1 } { 2 } a t ^ { 2 } + v _ { o } t + s _ { o }$, where $a$ is the acceleration, $v_o$ is the initial velocity, and $s_o$ is the initial position. We can use the given data to find the values of $v_o$ and $s_o$.At $t = 1$ second, $s = 234$ feet, so$$234 = \frac { 1 } { 2 } (-32) (1)^2 + v _ { o } (1) + s _ { o }$$$$234 = -16 + v _ { o } + s _ { o }$$$$v _ { o } + s _ { o } = 250$$At $t = 2$ seconds, $s = 202$ feet, so$$202 = \frac { 1 } { 2 } (-32) (2)^2 + v _ { o } (2) + s _ { o }$$$$202 = -64 + 2 v _ { o } + s _ { o }$$$$2 v _ { o } + s _ { o } = 266$$Subtracting the first equation from the second equation, we get$$v _ { o } = 16$$Substituting this value of $v_o$ into the first equation, we get$$s _ { o } = 234$$Therefore, the position equation is$$s = \frac { 1 } { 2 } (-32) t ^ { 2 } + 16 t + 234$$$$s = -16 t ^ { 2 } + 16 t + 234$$

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