# A car travel 45 mph at 9:00 pm, then another car leave 30 minutes later at 65 mph. When will they meet?

*print*Print*list*Cite

### 2 Answers

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.**

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.**