How to Solve a Catching-Up Problem

Someone set off ahead of you and you are chasing them. To find how long it takes to catch up, divide the head start by the difference between the two speeds.

$d$ is the head start, $a$ is the speed of the chaser, $b$ is the speed of the one ahead and $t$ is the time to catch up. The distance each covers, $at$ and $bt$, is shown as well.

Why the speeds are subtracted

Both move the same way, so the one ahead keeps going too. In every minute the gap closes only by what the chaser covers less what the one ahead covers, which is $a - b$. A closing-gap problem adds the speeds; this one subtracts them.

Example

Someone 600 m ahead walks at 80 m a minute and is chased at 200 m a minute. The gap closes by $200 - 80 = 120$ m a minute, so $t = 600 \div 120 = 5$ minutes. By then the chaser has covered $200 \times 5 = 1000$ m and the one ahead $80 \times 5 = 400$ m, a difference of exactly the original 600 m.

Notes

The chaser has to be the faster of the two. At equal speeds the gap holds, and a slower chaser falls further behind. Neither case is calculated.

Keep the units of speed and distance consistent. Speeds per minute give an answer in minutes.

This formula does not apply when the two move towards each other. Then the gap closes by the sum of the speeds, which is a closing-gap problem.