# A car travel 45 mph at 9:00 pm, then another car leave 30 minutes later at 65 mph. When will they meet? The first car leaves at 9:00 at 45 mph and 30 minutes later another car leaves at 65 mph. Let us assume they travel at a uniform rate throughout the journey.

In 30 minutes the first car has traveled 22.5 miles. The difference between the velocities of the two cars...

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The first car leaves at 9:00 at 45 mph and 30 minutes later another car leaves at 65 mph. Let us assume they travel at a uniform rate throughout the journey.

In 30 minutes the first car has traveled 22.5 miles. The difference between the velocities of the two cars is 65 - 45 = 20 mph.

The time taken by the second car to travel an extra 22.5 miles is 22.5/ 20 = 1.125 hours

1.125 hours  = 67.5 minutes or 1 hour 7.5 minutes.

So the cars will meet 1 hour 7.5 minutes after the first has left.

The cars meet at 7.5 minutes past 10.

Approved by eNotes Editorial Team Let the distance both cars travels before they meet is D.

The time need to for car 1 is T1

The car needed for car 2 is T2

But car 2 leaves 30 minutes ( 0.5 h) after car 1

==> T2 = T1 - 0.5 ..............(1)

Let us use the speed formula.

For the first car:

==> S1 = D/T1

==> 45 = D/T1

==> D= 45*T1..............(2)

For the second car:

s2 = D/T2

==> 65 = D/ (T1-0.5)

==> D= 65*(T1-0.5) ..............(3)

Now from (2) and (3) we have:

45*T1 = 65*(T1 - 0.5)

==> 45T1 = 65T1 - 65/2

==> 20T1 = 65/2

==> T1 = 65/2*20 = 13/8 hours= 1 5/8 hours

We will convert 5/8 hour to minutes.

==> 5/8 * 60 = 37.5 minutes

Then the time needed for car 1 to meet car 2 is 1:37.5

But car 1 leaves at 9:00 pm

==> 9:00 + 1:37.5 = 10:37.5

Then, the cars meets at 10:37.5 pm.

Approved by eNotes Editorial Team