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

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.