Question 195768
The equations for both hare and dog are:
{{{d[h] = r[h]*t[h]}}}
and
{{{d[d] = r[d]*t[d]}}}
These are laws of constant speed motion, and 
the key to not getting confused is to only
apply them when the hare and dog are moving.
I would like to have them both moving from 
start to finish, but how do I do that?
  I have a stopwatch and I will start it 
when the dog starts running. But where is the hare
then? The hare has a 1 minute headstart, and
in that time, she goes
{{{d[h] = 12*(1/60)}}}
{{{d[h] = .2}}} km
I start my stopwatch and stop it when the dog
catches the hare
{{{d[h] = r[h]*t[h]}}}
{{{d[h] = 12t[h]}}}
The dog has to make up 100 m plus 200 m
{{{d[d] = d[h] + .3}}} km
{{{d[h] + .3 = 16*t[d]}}}
The dog's time and the hare's time will now be the
same on my stopwatch, so I'll say they both are {{{t}}}
{{{d[h] + .3 = 16t}}}
(1) {{{d[h] = 16t - .3}}}
and
(2) {{{d[h] = 12t}}}
Subtract (2) from (1)
(1) {{{d[h] = 16t - .3}}}
(2) {{{-d[h] = -12t}}}
{{{0 = 4t - .3}}}
{{{4t = .3}}}
{{{t = .075}}} hr
{{{t = .075*60}}} min
{{{t = 4.5}}} min
It takes the dog 4.5 min to overtake the hare
How far does the hare go in that time?
(1) {{{d[h] = 16t - .3}}}
{{{d[h] = 16*.075 - .3}}}
{{{d[h] = 1.2 - .3}}}
{{{d[h] = .9}}} km
But the hare is already 200 m from her starting point, so
The dog overtakes the hare at
{{{.9 + .2 = 1.1}}} km 
1100 m from the hare's starting point
check answer:
The dog has to go 
{{{.1 + 1.1 = 1.2}}} km  in .075 hr to catch the hare
{{{d[d] = 16*t}}}
{{{1.2 = 16*.075}}}
{{{1.2 = 1.2}}}
OK
I don't have time for the other question, sorry