Answer:
a).The boat moved
x=1.16m
b). The speed canoe at this instant
v=1.3 m/s
Explanation:
Let consider a system of n' particles with respectives velocities and accelerations so for the distance
a)
Let the boat moves behind by x m from center of mass so
[tex]x_{t}=\frac{m_{p}*x'+m_{d}*x'+m_{b}*x'}{m{p}+m_{d}+m_{b}}[/tex]
[tex]x_{t}=\frac{95kg*\frac{3m}{2}'+25kg*\frac{3m}{2}'+62kg*0'}{95kg+25kg+60kg}[/tex]
[tex]=\frac{95kg*(x-1.5)-25kg*(1.5+x)+62kg*(-x)}{95kg+25kg+60kg}[/tex]
Resolve to x'
[tex]180x+105=-105[/tex]
[tex]x=1.16[/tex]
b).
Vcom=0
[tex]v_{t}=\frac{m_{p}*v'+m_{d}*x'+m_{b}*v'}{m{p}+m_{d}+m_{b}}[/tex]
[tex]x_{t}=\frac{95*(1.3)'+25*(1.3)'+60*V'}{m{p}+m_{d}+m_{b}}[/tex]
Resolve to v'
[tex]78-60v=\\[/tex]
[tex]v=\frac{78}{60}[/tex]
[tex]v=1.3 \frac{m}{s}[/tex]
