SOLUTION: Please help me i have no clue how to do this! Years after the famous foot race, the hare has challenged the turtle to a rematch on bycicles. that hare can pedal a rate of 15mph

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Question 128016: Please help me i have no clue how to do this!
Years after the famous foot race, the hare has challenged the turtle to a rematch on bycicles. that hare can pedal a rate of 15mph while the tortise can only travel 12 mph. knowing he was slower, the tortoise requested haveing a 27 mile head start in the 165 mile race.
1. WHo won the race? by How much Time?
2. How far into the Race did the hare overtake the tortise?
3. What is the largest head start the tortise could have been given without being able to beet the hare?
4. suppose a chipmunk entered the race with a 84 miles head start and can pedal at 9 mph. develope an equation for the chipmunk and calculate when the hare would finally catch up to the chipmunk assuming the race doesn't end @ 165 miles.
5.
a pig enters the race. he starts at the finish line riding backwards, at a rate of 15mph where and when will he meet the hare on the road?
THank you

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Start your stopwatch when the hare starts
the race. 165+-+27+=+138 is the distance
the tortoise has left to go.
His time is 138+%2F+12+=+11.5 hrs to the
finish line. The hares time on the same stopwatch
is 165+%2F+15+=+11hrs, so the hare gets to
the finish 1/2 hr before the tortoise
----------------------------------------
When the stopwatch starts, when has the tortoise
gone d mi and the hare gone d+%2B+27mi
Their times will be the same.
t+=+d+%2F+12 for the tortoise
t+=+%28d+%2B+27%29+%2F+15 for the hare
d%2F12+=+%28d+%2B+27%29+%2F+15
multiply both sides by 12%2A15
15d+=+12%28d+%2B+27%29
15d+=+12d+%2B+324
3d+=+324
d+=+108mi
So, when the tortoise has gone 108 mi and the
hare has gone 108+%2B+27+=+135 mi, the hare
passes the tortoise. This is t+=+108%2F12
or t+=+9 hrs on the stopwatch
--------------------------------
t+=+d+%2F+12 for the tortoise
t+=+%28d+%2B+h%29+%2F+15 for the hare
h is the headstart for the tortoise.
Make their times equal on the stopwatch and
set d+=+165+-+27 or d+=+138
138%2F12+=+%28138+%2B+h%29+%2F+15
15%2A138+=+12%28138+%2B+h%29
2070+=+1656+%2B+12h
12h+=+2070+-+1656
12h+=+414
h+=+34.5 mi
If the totoise gets a 34.5 mi headstart,
they will meet at the finish line
-----------------------------------
For chipmunk:
t+=+d+%2F+9
For hare:
t+=+%28d+%2B+84%29+%2F+15
set the times equal
d++%2F9+=+%28d+%2B+84%29+%2F+15
15d+=+9%2A%28d+%2B+84%29
15d+=+9d+%2B+756
6d+=+756
d+=+126
The chipmunk has gone 126 mi and
the hare has gone d+%2B+84+=+210 mi
The time is t+=+d%2F+9 or t+=+126+%2F+9
t+=+14 hrs on the stopwatch
------------------------------------
The pig and the hare are heading towards
eachother at the same rate.
When they meet, their times will be the
same on the stopwatch, and the sum of the
distances they cover will be 165mi.
For hare:
t+=+d%5Bh%5D+%2F+15
For pig:
t+=+d%5Bp%5D+%2F+15
Their times are equal, so
d%5Bh%5D+%2F+15+=+d%5Bp%5D+%2F+15
d%5Bp%5D+=+d%5Bh%5D
and
d%5Bp%5D+%2B+d%5Bh%5D+=+165
df%5Bp%5D+=+82.5
d%5Bh%5D+=+82.5
t+=+82.5+%2F+15
t+=+5.5 hrs
The pig and the hare will meet
in 5.5 hours halfway to the finish