Hello, asnxbbyx113!

Another approach to #2 . . .

2) A man starts walking north at 5ft/s from a point $\displaystyle P.$ .Five minutes later,

a woman starts walking south at 4ft/s from a point $\displaystyle Q$ 1300 ft due east of $\displaystyle P.$

At what rate are the people moving apart 20 min after the woman starts walking?

The man walks north from $\displaystyle P$ at 5 ft/sec.

Since he has a 5-minute (300-second) headstart.

. . he has already walked: $\displaystyle 5 \times 300 \:=\:1500$ feet.

In the next $\displaystyle t$ seconds, he walks $\displaystyle 5t$ feet more to $\displaystyle A.$

The woman walks south from $\displaystyle Q$ at 4 ft/s.

In the next $\displaystyle t$ seconds, she walks $\displaystyle 4t$ feet to $\displaystyle B.$

The diagram looks like this: Code:

A *
| *
5t| *
| *
+ *
| * R
1500| *
| *
P *---------------*-------* Q
: * |
4t: * |4t
: * |
C + - - - - - - - - - - - * B
: 1300 :

The distance between them is: $\displaystyle R \,=\,AB.$

From the right triangle $\displaystyle ACB$, we have:

. . $\displaystyle R^2 \:=\:(5t + 1500 + 4t)^2 + 1300^2\quad\Rightarrow\quad R \:=\:\left[(9t + 1500)^2 + 1300^2\right]^{\frac{1}{2}}$

Differentiate: .$\displaystyle \frac{dR}{dt} \:=\:\frac{1}{2}\left[(9t + 1500)^2 + 1300^2\right]^{-\frac{1}{2}}\left[2(9t + 1500)\cdot9\right] $

. . and we have: .$\displaystyle \frac{dR}{dt} \:=\:\frac{9(9y + 1500)}{\sqrt{(9t + 1500)^2 + 1300^2}} $

"20 minutes after the woman starts walking" means: $\displaystyle t = 1200$ seconds.

. . Plug it in and simplify . . .