Question

A tortoise crawling at a rate of .1mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at a rate of .5mi/h. How many feet must the hare run to catch the tortoise?

Answer 1

we start with an easy question
in 30 minutes is what portion of an hour?

Answer 2

half

Answer 3

k good
so in half an hour at a rate of \(.1\) mph how far did the tortoise go?

Answer 4

In ft, he went 264ft

Answer 5

lets keep it in miles, since that is the unit given
other wise we are going to get all messed up with the units

Answer 6

Well, then he would have traveled .05 miles

Answer 7

k good
now we use "distance equals rate times time"

Answer 8

the tortoise has rate \(.1\) and a \(.05\) mile head start, so his distance is going to be
\[.05+.1t\] where \(t\) is time (in hours)

Answer 9

the hare has rate \(.5\) so his distance is going to be \(.5t\)

Answer 10

and since evidently they meet at the same distance, you can put
\[.05+.1t=.5t\] and solve for \(t\)

Answer 11

t= -1?

Answer 12

not too hard with these decimals, or we can work with whole numbers by writing
\[5+10t=50t\] either way, you get
\[5=40t\] so
\[t=\frac{5}{40}=\frac{1}{8}\]

Answer 13

or
\[.05+.1t=.5t\\.05=.4t\\
\frac{.05}{.4}=t\]still get \(t=\frac{5}{40}=\frac{1}{8}\)

Answer 14

or if you like decimals \(t=.125\)
now we find the distance
it is \(.5\times .125=.0625\) miles
convert to feet by multiplying
\[5280\times .0625\]

Answer 15

So, the hare has to travel 330 ft?