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, $s = 273$ feet
At $t = 2$ seconds, $s = 241$ feet
At $t = 3$ seconds, $s = 177$ feet
A) $s = - 32 t ^ { 2 } + 16 t + 273$
B) $s = - 16 t ^ { 2 } + 16 t - 273$
C) $s = - 16 t ^ { 2 } + 16 t + 273$
D) $s = - 8 t ^ { 2 } + t - 273$
E) $s = - 16 t ^ { 2 } - 16 t + 273$

Question

Quizplus · Accepted Answer

Rearranging the position equation, we have $$s=\frac{1}{2}at^2+v_0t+s_0$$ To solve for the coefficients, we can use the information given. Plugging in the first set of values, we get the equation $$273=\frac{1}{2}a(1)^2+v_0(1)+s_0$$ which simplifies to $$\frac{1}{2}a+v_0+s_0=273$$ Plugging in the second set of values, we get the equation $$241=\frac{1}{2}a(2)^2+v_0(2)+s_0$$ which simplifies to $$2a+2v_0+s_0=241$$ Plugging in the third set of values, we get the equation $$177=\frac{1}{2}a(3)^2+v_0(3)+s_0$$ which simplifies to $$\frac{9}{2}a+3v_0+s_0=177$$ We can now solve for the coefficients by setting up a system of equations. Subtracting the first equation from the second, we get $$\frac{3}{2}a+2v_0=-32$$ Subtracting the second equation from the third, we get $$\frac{5}{2}a+v_0=-64$$ Solving this system of equations, we get $$a=-32\text{ ft/sec}^2$$ $$v_0=48\text{ ft/sec}$$ Plugging in these values to any of the three equations, we can solve for $s_0$ and get $$s_0=273$$ Therefore, the position equation is $$s=-16t^2+16t+273$$ and the answer is C.

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