Answer:
They are 28.87m appart from each other 4s later. The distance between them is changing at a rate of 1.81m/s at that moment
Explanation:
In order to solve this problem, we must first do a drawing that will represent the situation. (See attached picture)
Notice that both Jamal and Brook are moving either on the east direction or the west direction. We are going to call these directions x.
Since they are moving relative to each other, we can find the relative x velocity between them (or how fast they are moving relative to each other in the x-direction). We can do this by adding the two velocities, like this:
[tex]V_{r}=V_{J}+V_{B}[/tex]
so
[tex]V_{r}=7km/hr+6km/hr=13km/hr[/tex]
now, for it to be easier for us to calculate, we can turn that to m/s, so we get:
[tex]13km/hr*\frac{1000m}{1km}*\frac{1hr}{3600s}=3.611m/s[/tex]
once we did this conversion we can find out how far appart they are to each other in the x-direction by using the following formula:
[tex]V=\frac{x}{t}[/tex]
we can solve this for x so we get:
x=Vt
when substituting the values we get:
x=(3.611m/s)(4s)=14.44m
with this information we can draw a triangle that will represent the distance between them. (See second diagram)
As we can appreciate on the second diagram, we can model the situation as a right triangle, which can be solved by using pythagore:
[tex]d^{2}=x^{2}+y^{2}[/tex]
we can solve this for d so we get:
[tex]d=\sqrt{x^{2}+y^{2}}[/tex]
we can now substitute the correspoding data so we get:
[tex]d=\sqrt{(14.44m)^{2}+(25m)^{2}}[/tex]
which yields:
d=28.87m
Once we got our first answer, we can use it to find how fast they are moving away from each other at that precise moment.
we take pythagore again so we get:
[tex]d^{2}=x^{2}+y^{2}[/tex]
we know that y will always be constant so we can substitute it now:
[tex]d^{2}=x^{2}+(25)^{2}[/tex]
which yields:
[tex]d^{2}=x^{2}+625[/tex]
since x will change over time we will not substitute it yet.
Now we can use implicit differenciation to find the velocity we want:
2dd'=2xx'
in this case d' represents the velocity at which they are moving appart from each other and x' represents the x velocity they will have relative to each other, so we can rewrite the equation as:
[tex]2dV_{d}=2xV_{x}[/tex]
we can now sove this for [tex]V_{d}[/tex] so we get:
[tex]V_{d}=\frac{x}{d}V_{x}[/tex]
so we can now substitute the values we know:
[tex]V_{d}=\frac{14.44m}{28.87m}(3.611m/s)[/tex]
so
[tex]V_{d}=1.81m/s[/tex]
The distance between the two parallel paths is changing at that moment at 1.8 m/s.
Let (x,0) represent Brooke distance at tome t and (y, 25) represent Jamail distance at time t.
Since Brooke jogs at 6 km/h(1.67 m/s), hence 1.67 = x/t; x = 1.67t
Since Jamail jogs at 7 km/h(1.94 m/s) in opposite direction(-ve), hence -1.94 = y/t; y = -1.95t
The distance (D) between Brooke and Jamail at time t is:
[tex]D=\sqrt{(x-y)^2+(0-25)^2} \\\\D=\sqrt{(x-y)^2+625}\\\\D^2={(x-y)^2+625}\\\\D^2=(1.67t-(-1.94t))^2+625\\\\D^2=13t^2+625\\\\Also:\\\\2D\frac{dD}{dt}= 26t\\\\\frac{dD}{dt}=\frac{13t}{D}\\\\\\Hence, at\ t=4:\\\\D^2=13(4)^2+625=833\\\\D=28.86m\\\\\\\frac{dD}{dt}=\frac{13(4)}{28.86}=1.8\ m/s[/tex]
The distance between the two parallel paths is changing at that moment at 1.8 m/s.
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