The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.

<h3>Separation distance</h3>

Jason got a head start of 20 miles, so that is the initial separation between the two drivers.

<h3>Closure speed</h3>

Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.

<h3>Closure time</h3>

The relevant relation is ...

time = distance/speed

Then the time it takes to reduce the separation distance to zero is ...

closure time = separation distance / closure speed = 20 mi / (10 mi/h)

closure time = 2 h

Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.

<em>Additional comment</em>

The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.