Mike and Ike decide to do a little racing, but Ike is slower, so he's given a head start. Ike passes the one-mile mark doing 60 miles per hour! 1 minute later, Mike passes the same point traveling 71 miles per hour. How many minutes after the start of the race will Mike catch up to Ike, assuming Mike and Ike travel at constant speeds?

 May 4, 2022

Let t minutes be the time needed.


Mike is traveling at \(\dfrac{71}{60}\text{ mi/min}\) and Ike is traveling at \(1\text{ mi/min}\).

1 minute after Ike passed the one-mile mark, Mike passes the one-mile mark. But in that one minute that Mike used to catch up to Ike, Ike has traveled an extra 1 mile. So at that specific moment:



            1 mi            Mike        1 mi            Ike

t minutes after that moment, total distance travelled by Mike is \(\left(1 + \dfrac{71}{60}t\right)\) miles and total distance travelled by Ike is \((2 + t)\) miles.


When Mike catches up to Ike, the total distance traveled will become the same. so \(1 + \dfrac{71}{60}t = 2+ t\). Please solve for t on your own.

Then (t + 2) minutes is the answer.

 May 4, 2022

Thank you so much!

 May 5, 2022

12 Online Users