Exercise $2\mathrm{.}50$: A small object moves along the $x-$axis with acceleration $a_x(t)=-(0\mathrm{.}0320\frac{(m)}{s^3})(15\mathrm{.}0s-t).$ At $t=0$ the object is at $x=-14\mathrm{.}0m$ and has velocity $v_{0x}=8\mathrm{.}00\frac{(m)}{(s)}.$ What is the $x-$coordinate of the object when $t=10\mathrm{.}0s?$

Identify: The acceleration is not constant, so we must use calculus instead of the standard kinematics formulas.

SET UP: The general calculus formulas are $v_2=v_{0,}{\displaystyle \underset{0}{\overset{\text{'}}{}}}a,dt$ and $x=x_0{\displaystyle \underset{0}{\overset{\text{'}}{}}}{v}_{2}dt.$ First integrate $a_x$ to find $v(f),$ and then integrate that to find $x(f).$

EXECUTE: Find $v(t)$:$v_x(t)=v_{0,t}{\displaystyle \underset{0}{\overset{\text{'}}{}}}a,dt=v_{0,t}{\displaystyle \underset{0}{\overset{\text{'}}{}}}-(0\mathrm{.}0320\frac{m}{s^{\text{'}}})(15\mathrm{.}0s-t)dt.$ Carrying out the integral and putting in the numbers gives $[$:$\frac{t^2}{2}$