Average speed of slower car = 60 mph
Average speed of faster car = 70 mph
<em><u>Solution:</u></em>
Given that Two cars enter the Delaware turnpike at shoreline drive at 8 am, each heading for ocean city
Given that One cars average speed is 10 mph faster than the other
Let "x" be the average speed of slower car
Then x + 10 is the average speed of faster car
The faster car arrives at ocean city at 11 am, a half hour before the slower car
<em><u>Time taken by faster car:</u></em>
The faster car arrives at ocean city at 11 am but given that they start at 8 a.m
So time taken by faster car = 11 am - 8am = 3 hours
<em><u>Time taken by slower car:</u></em>
The faster car arrives at ocean city at 11 am, a half hour before the slower car
So slower car takes half an hour more than faster car
Time taken by slower car = 3 hour + half an hour = ![3\frac{1}{2} \text{ hour}](https://tex.z-dn.net/?f=3%5Cfrac%7B1%7D%7B2%7D%20%5Ctext%7B%20hour%7D)
Now the distance between Delaware turnpike at shoreline drive and ocean city will be same for both cars
Let us equate the distance and find value of "x"
<em><u>The distance is given by formula:</u></em>
![distance = speed \times time](https://tex.z-dn.net/?f=distance%20%3D%20speed%20%5Ctimes%20time)
<em><u>Distance covered by faster car:</u></em>
![distance = (x + 10) \times 3 = 3x + 30](https://tex.z-dn.net/?f=distance%20%3D%20%28x%20%2B%2010%29%20%5Ctimes%203%20%3D%203x%20%2B%2030)
<em><u>Distance covered by slower car:</u></em>
![distance = x \times 3\frac{1}{2} = x \times \frac{7}{2} = 3.5x](https://tex.z-dn.net/?f=distance%20%3D%20x%20%5Ctimes%203%5Cfrac%7B1%7D%7B2%7D%20%3D%20x%20%5Ctimes%20%5Cfrac%7B7%7D%7B2%7D%20%3D%203.5x)
Equating both the distance,
3x + 30 = 3.5x
3x - 3.5x = -30
-0.5x = -30
x = 60
Thus average speed of slower car = 60 mph
Average speed of faster car = x + 10 = 60 + 10 = 70 mph