Answer:
After 3 hours.
Explanation:
This is a encounter problem.
We start by writing the position equation of both objects.
We assume that acceleration is equal to zero in both boats ⇒ They move with constant speed
We put the origin of coordinates in the harbour so after 6 hours the first boat travels [tex]7\frac{mi}{h}.6h=42mi[/tex]
The first boat will be 42 mi far from the harbour.
The position equation for a motion with acceleration equal to zero is :
[tex]x(t)=x_{0}+v(t-t_{i})[/tex]
Where [tex]x_{0}[/tex] is initial position
v is the speed
t is time and [tex]t_{i}[/tex] is initial time
Then [tex](t-t_{i})[/tex] is time variation
For the first boat :
[tex]x1(t)=42mi+7\frac{mi}{h}.t[/tex]
Because we set [tex]t_{i}=0 h[/tex]
For the second one :
[tex]x2(t)=21\frac{mi}{h}.t[/tex]
In the encounter : x1(t) = x2(t) ⇒
[tex]42mi+7\frac{mi}{h}.t=21\frac{mi}{h}.t[/tex]
[tex]14\frac{mi}{h}.t=42mi\\ t=\frac{42}{14}.h\\ t=3h[/tex]
So after 3 hours they have the same position relative to the harbour
We can check by replacing the value t = 3h in both position equations
For x1(t) :
[tex]x1(3h)=42mi+7\frac{mi}{h}.(3h) \\x1(3h)=63mi[/tex]
For x2(t) :
[tex]x2(3h)=21\frac{mi}{h}.(3h)\\x2(3h)=63mi[/tex]
