Question: If you are to travel exactly 2 miles and your speed over the
first mile is 30.1 mph, what must your speed be over the second mile
to average 60 mph over the entire 2 miles?
You must prove your answer.

Are you sure it is 30.1 and not 30 ?

Its been a long time since I left school but let me give a shot:

Speed(s) = distance(d) / time(t)

In both the cases you would travel the same distance so use d=s*t

In first case,?you get 60*2 = 120

In second case you have 1*30.1+1*x (assuming the speed is x)

so we have 
120 = 1*30.1 + x
x=120-30.1
x=89.9

So is 89.9 mph the correct answer ?

It takes two minutes to travel two miles at an average of 60
miles/hour (mph).  To travel one mile at 30 mph would also take two
minutes leaving no time at all to travel the second mile.  So it can't
be 30 mph for the first mile.  It must be 30.1 mph.  Clearly very
little time is left for the second mile.

The  formula is:  Distance(d) = Speed(s) x Time(t)

We will express time in hours, not minutes since we are interested in
the speed expressed in miles/hour.

(1) Calculate the amount of time to travel the first mile at 30.1 mph:

t = d/s = 1 mile / 30.1 mile/hour = 1 hour / 30.1

(2) Time available to travel the second mile

t2 = (2/60)hour - (1 hour / 30.1) = (1 hour / 30) - (1 hour / 30.1)

(3) Speed for second mile

s2 = d/t2 = 1 mile / ((1 hour / 30) - (1 hour / 30.1)) = 9030 mile/hour

Wow, 3rd grade flashback to where I learned about division by zero.
The original problem-- one of theose extra credit ones-- was "first
mile at 30 mph" and the solution was that the problem was unsolvable,
as Ansel001 mentions.

Looks like teachers got tired of making the kids nuts. Must've been
bad for their psyche or something. But I turned out fine. After all, I
have the heart of a Russian soldier, the brain of a physicist, and the
body of a lingerie model.

They're in the trunk of my car.

Hi,I'm afraid the first "answer" is not correct,but the second is right.
Let's prove it.
1.we caculate the total time
 total time = total distance / average speed 
            =       2        /     60
            =   1/30        
2.we caculate the first 1 mile's time
       time = distance / average speed
            =     1    /    30.1
            =   1/30.1
3.we caculate the last 1 mile's time
  this time = total time - the time above that we just caculated
            =   1/30     -           1/30.1
            =(1/30-1/30.1)
4.we caculate the last 1 mile's speed
 this speed = distance  / this time
            =    1      /  (1/30-1/30.1)
            =9030
so this is the correct speed during the second mile.
remember the formula:
       average speed= distance / time
let's then see how the first "answer" is not correct.

he wrote:   "In first case,you get 60*2=120"
   this is wrong.....60,which is a speed,but what does the number "2"
mean? time? no,we are not provided with this information;the total
distance 2 miles?no,what should this speed*distance be......we do not
have this formula....it's nonsense....
   that's all,if you have anyother problems you can ask me,thank you~~

by  Cecil

The answer is simple if you read the question
I translate you travel 2 miles the first mile with a speed of 30.1 mph
and the second mile with an unknown speed. But you know that the
average speed over two miles is 60 mph.
The answer is 60= (30.1+ X)/2 => 120 = 30.1 + X => 89.9 mph
Proof (30.1+ 89.9)/2 =....= 60 mph
Do you have a more difficult one???
Prof. Kees
So jaspreet123-ga is the one with the good answer but he forgot the proof

Professor,

You are mistaken.

You are not traveling at the two speeds of 30.1 mph and x mph for the
same amount of time so you can't just average them to 60.  It takes 1
minute, 59.6 seconds to travel the first mile at 30.1 mph.  You will
be traveling for less than half a second at the second speed for the
last mile.  My answer of 9030 mph for the second mile is correct.

I always liked these questions where if you think about the answer in
the context actual situation it is just ridiculous, like even if you
were to have means to get in a car that could go from 30.1 mph
uniformly up to some speed x mph in that fraction of a second you
would be mush, I know it is to help students but it always made me
chuckle about the poor sap who experienced the 26846 g's (at least I
think that is it, haven't slept in a few days so that could be off, or
who knows maybe spark another 9030 v 89.9 debate).

9030mph is the correct answer for an AVERAGE velocity for the second
leg of the journey. The problem can be idealized to two velocities,
two legs... here's the painfully "proven" answer - where all equations
refer to the definition of velocity or the simple additive property of
times and distances. (obviously, this is a non-relativistic situation,
since it was never specified in what reference frame the times or
distances are measured).

Working equations:  where I use 1 for the first leg of the trip, and 2
for the second:

1) v1=d1/t1 (velocity definition, first leg)
2) v2=d2/t2 (velocity definition, second leg)
3) v=d/t    (velocity definition, overall)
4) d1+d2=d  (distances are additive)
5) t1+t2=t  (times are additive)

Givens:
d=2mi, d1=1mi, d2=1mi
v1=30.1mph, v=60mph
v2, t1, t2, t are unknown

t1 = d1/v1 = 1mi/30.1mph = 1hr/30.1 = (10/301)hr  (equation 1)
t = d/v = 2mi/60mph = 1hr/30 = (1/30)hr  (equation 3)
t2 = t - t1 = (1/30)hr-(10/301)hr = (301 - 300)/9030 hr = (1/9030)hr   (equation 5)

The average speed in the second leg will have to be:
v2 = 1mi/(1/9030)hr = 9030mph...  or about mach 11.7

recheck:
1) 30.1mph=1mi/(10/301)hr check
2) 9030mph=1mi/(1/9030)hr check
3) 60mph  =2mi/(1/30)  hr check
4) 1mi + 1mi = 2mi        check
5) (10/301)hr + (1/9030)hr = (301/9030)hr = (1/30)hr check

Looks like the best professor needs a new moniker.

The formula for this is   X * Y  / 2X - Y  
Where X is the first mile speed 
and  Y is the overall speed.

The answer is 9030 miles/hour.