Tortiose racing

Two tortoises named A and B must run a race. A starts with an average speed of 720 feet per hour. Young B knows she runs faster than A, and furthermore has not finished her cabbage.

When she starts, at last, she can see that A has a 70 feet lead but B's speed is 850 feet per hour. How long will it take B to catch A?

More generally: given two speeds v1 (A's speed, integer > 0) and v2 (B's speed, integer > 0) and a lead g (integer > 0) how long will it take B to catch A?

The result will be an array [hour, min, sec] which is the time needed in hours, minutes and seconds (round down to the nearest second) or a string in some languages.

If v1 >= v2 then return nil, nothing, null, None or {-1, -1, -1} for C++, C, Go, Nim, Pascal, COBOL, Erlang, [-1, -1, -1] for Perl,[] for Kotlin or "-1 -1 -1".

Examples:

(form of the result depends on the language)

race(720, 850, 70) => [0, 32, 18] or "0 32 18"
race(80, 91, 37)   => [3, 21, 49] or "3 21 49"

Note:

• See other examples in "Your test cases".
• In Fortran - as in any other language - the returned string is not permitted to contain any redundant trailing whitespace: you can use dynamically allocated character strings.

** Hints for people who don't know how to convert to hours, minutes, seconds:

• Tortoises don't care about fractions of seconds
• Think of calculation by hand using only integers (in your code use or simulate integer division)
• or Google: "convert decimal time to hours minutes seconds"

---

import math
def race(v1, v2, g):
    alist=[]
    temp=0
    if v2<=v1:
        return None
    sec=(g/((v2-v1)/3600))
    alist.append(sec//3600)
    alist.append((sec%3600)//60)
    alist.append(math.floor((sec%3600)%60))
    return alist