Question

Two cars enter the Delaware turnpike at shoreline drive at 8 am, each heading for ocean city. One cars average speed is 10 mph faster than the other. The faster car arrives at ocean city at 11 am, a half hour before the slower car. What is the average speed of each car?

Answer 1

Average speed of slower car = 60 mph

Average speed of faster car = 70 mph

Solution:

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

Time taken by faster car:

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

Time taken by slower car:

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 = $3\frac{1}{2} \text{ hour}$

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"

The distance is given by formula:

$distance = speed \times time$

Distance covered by faster car:

$distance = (x + 10) \times 3 = 3x + 30$

Distance covered by slower car:

$distance = x \times 3\frac{1}{2} = x \times \frac{7}{2} = 3.5x$

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