Solve the problem.
-Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time $t$, fint body's position at time $t$. $a = 3.2 , v ( 0 ) = - 14 , s ( 0 ) = - 5$
A) $\mathrm { s } = 1.6 \mathrm { t } ^ { 2 } - 14 \mathrm { t }$
B) $s = - 1.6 t ^ { 2 } + 14 t - 5$
C) $s = 3.2 t ^ { 2 } - 14 t - 5$
D) $\mathrm { s } = 1.6 \mathrm { t } ^ { 2 } - 14 \mathrm { t } - 5$

Question

Quizplus · Accepted Answer

The position $s$ at time $t$ can be found using the equation $s = s_0 + v_0t + \frac{1}{2}at^2$, where $s_0$ is the initial position, $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Substituting the given values ($s_0 = -5$, $v_0 = -14$, $a = 3.2$) into the equation gives $s = -5 - 14t + \frac{1}{2}(3.2)t^2$, which simplifies to $s = 1.6t^2 - 14t - 5$.

Solve the Problem ttt , Fint Body's Position at Time ttt a=3.2,v(0)=−14,s(0)=−5a = 3.2 , v ( 0 ) = - 14 , s ( 0 ) = - 5a=3.2,v(0)=−14,s(0)=−5

Solve the Problem $t$ , Fint Body's Position at Time $t$ $a = 3.2 , v ( 0 ) = - 14 , s ( 0 ) = - 5$