Answer
The average speed for the slower car was 40 miles per hour.
The average speed for the faster car was 50 miles per hour.
Both traveled the same distance of 100 miles.
Work Step by Step
We'll say that v1 is the speed for the slower car and v2 for the faster car, so:
$v_{2}=v_1+10$
We know that v=d/t and the distance the cars traveled were the same, so:
$\frac{d}{t_2}=\frac{d}{t_1}+10$
The time it took for the faster car to arrive at Wildwood was 2 hours and 2.5 hours for the other one, so:
$\frac{d}{2}=\frac{d}{2.5}+10$
Now, we can solve for d to find the distance traveled:
$\frac{d}{2}\cdot10=(\frac{d}{2.5}+10)\cdot10$
$5d=4d+100$
$5d-4d=4d+100-4d$
$d=100$
The distance traveled was 100 miles.
The average speed in miles per hour for each car is:
$v_1=100/2.5=40$
$v_{2}=v_1+10=40+10=50$
