Answer:
11.98 m
Explanation:
Given that:
mass of the child [tex]m_c[/tex] = 53 kg
mass of the boat [tex]m_b[/tex] = 107 kg
[tex]\text{length of the boat L = 7 m}[/tex]
the distance of the boat from pies l = 7.3 m
initial momentum [tex]P_i = 0[/tex]
Final momentum [tex]P_f = mc \dfrac{L}{f}- (m_c +m_b) \dfrac{x}{l}[/tex]
where;
x = distance moved by boat towards left
t = time taken for the child to travel to the far end of the boat
[tex]P_i =P_f[/tex]
∴
[tex]m_c \dfrac{L}{t}=(m_c +m_b) \dfrac{x}{t}[/tex]
Then;
[tex]x = \dfrac{m_cL}{m_c+m_b}[/tex]
[tex]x = \dfrac{53 \times 7}{53+107}[/tex]
x = 2.32 m
The distance of the child from the pier is:
d = L +(l - x)
d = 7 m + ( 7.3 m - 2.32 m)
d = 7 m + 4.98 m
d = 11.98 m