Let's first find acceleration from the Newtons Second Law of Motion:
"F_{appl}-F_{fr}=(m_{cart}+m_{child})a,""F_{appl}-\\mu_k (m_{cart}+m_{child})g=(m_{cart}+m_{child})a,""a=\\dfrac{F_{appl}-\\mu_k (m_{cart}+m_{child})g}{(m_{cart}+m_{child})},""a=\\dfrac{100\\ N-0.2\\cdot (30\\ kg+\\dfrac{30\\ N}{9.8\\ \\dfrac{m}{s^2}})\\cdot 9.8\\ \\dfrac{m}{s^2}}{(30\\ kg+\\dfrac{30\\ N}{9.8\\ \\dfrac{m}{s^2}})}=1.1\\ \\dfrac{m}{s^2}."Finally, we can find the distance traveled by the cart in 3 s from the kinematic equation:
"d=v_0t+\\dfrac{1}{2}at^2=0+\\dfrac{1}{2}\\cdot 1.1\\ \\dfrac{m}{s^2}\\cdot (3\\ s)^2=4.95\\ m."Answer:
"d=4.95\\ m."
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