Question

two nascar cars are racing around a 2 mile track. one car can complete a lap in 48 seconds, and the other car can complete one lap in 40 seconds. if they start side by side, after how much time will they be side by side again

Answer 1

Let's assume that the faster car has a 2 mile "head start".
The faster car is 1.2 times faster (48/40) than the slower car.
It will take five 2 mile trips to "catch" the other car.
Five 2 mile trips will take 5 * 40 seconds = 200 seconds.

Answer 2

First we need to find the speed of each car using the formula

$s=\frac{d}{t}$, where s is the speed, d is the distance and t is time.

$s=\frac{2}{48}=0.0417 miles/s$

$s=\frac{2}{40} = 0.05miles/s$

Since they are moving in the same direction we subtract the speeds to obtain relative velocity, $V_r$

$V_r = 0.05-0.0417= 0.0083 miles/s$, then

$t=\frac{2}{0.0083}= 240s$.  This is the time it takes for the fastest car to complete an extra lap relative to the slower car.