Respuesta :
[tex]\begin{gathered} \text{For car} \\ v_1=11.6\text{ m/s} \\ \Delta x=0.287m \\ v_2=\text{ 0 m/s} \\ t=\text{?} \\ To\text{ find t} \\ v^2_2=v^2_1+2a\Delta x \\ \text{Solving a} \\ v^2_2-v^2_1=2a\Delta x \\ a=\frac{v^2_2-v^2_1}{2\Delta x} \\ a=\frac{(0m/s)^2-(11.6m/s)^2}{2(0.287m)} \\ a=\frac{0m^2/s^2-134.56m^2/s^2}{0.574m} \\ a=\frac{-134.56m^2/s^2}{0.574m} \\ a=-234.42m/s^2 \\ \text{Then} \\ t=\frac{v_2-v_1}{a} \\ t=\frac{0\text{ m/s-11.6m/s}}{-234.42m/s^2} \\ t=\frac{\text{-11.6m/s}}{-234.42m/s^2} \\ t=0.0495\text{ s} \\ \text{For child} \\ m=21.2\text{ kg} \\ v_1=11.6\text{ m/s} \\ v_2=\text{ 0 m/s} \\ t=0.0495\text{ s} \\ F=\text{?} \\ F=\frac{P_2-P_1}{t} \\ P_2-P_1=mv_2-mv_1=m(v_2-v_1),\text{ then} \\ F=\frac{m(v_2-v_1)}{t} \\ F=\frac{(21.2kg)(0\text{ m/s-11.6m/s})}{0.0495\text{ s}} \\ F=\frac{(21.2kg)(-11.6\text{ m/s})}{0.0495\text{ s}} \\ F=\frac{-245.92\operatorname{kg}\text{ m/s}}{0.0495\text{ s}} \\ F=-4968\text{ N} \\ \text{The force to stop the child is 4968 N. The negative means } \\ \text{the force is opposite to the movement} \end{gathered}[/tex]
