Find the position equation $s = \frac { 1 } { 2 } a t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ for an object that has distance $s = 24 \text { feet }$ at t=1 second, $s = 47 \text { feet }$ at t=2 seconds, and $s = 78 \mathrm { feet }$ at $t = 3$ seconds.
A) $s = 9 t ^ { 2 } + 4 t + 11$
B) $s = 4 t ^ { 2 } + 11 t + 9$
C) $s = 3 t ^ { 2 } + 9 t + 4$
D) $s = 11 t ^ { 2 } + 4 t + 9$
E) $s = 4 t ^ { 2 } + 3 t + 11$

Question

Quizplus · Accepted Answer

To find the correct equation, we substitute the given values of $s$ and $t$ into the position equation $s = \frac { 1 } { 2 } a t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ and solve for $a$, $v_0$, and $s_0$. Using the given points:1. At $t=1$, $s=24$: $24 = \frac{1}{2}a(1)^2 + v_0(1) + s_0$2. At $t=2$, $s=47$: $47 = \frac{1}{2}a(2)^2 + v_0(2) + s_0$3. At $t=3$, $s=78$: $78 = \frac{1}{2}a(3)^2 + v_0(3) + s_0$Solving these equations simultaneously for $a$, $v_0$, and $s_0$ will give us the values that fit one of the given options. The correct values that fit these conditions are $a = 8$, $v_0 = 11$, and $s_0 = 9$, which corresponds to the equation $s = 4t^2 + 11t + 9$, making option B the correct choice.

Find the Position Equation s=12at2+v0t+s0s = \frac { 1 } { 2 } a t ^ { 2 } + v _ { 0 } t + s _ { 0 }s=21​at2+v0​t+s0​

Find the Position Equation $s = \frac { 1 } { 2 } a t ^ { 2 } + v _ { 0 } t + s _ { 0 }$