At 11am John starts along a highway at 50mph then 15 minutes later,Jimmy drives same highway at 65mph when will Jimmy catch up with John?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

John starts along the highway at 50 mph at 11:00, Jimmy starts 15 minutes later or at 11:15 with a speed of 65 mph. I assume they move at a constant speed once they have started.

For Jimmy to catch up with John he has to cover the extra distance that John has travelled as he started 15 minutes earlier. In 15 minutes John moves 50*(15/60) = 12.5 miles.

Jimmy is moving 65 - 50 or 15 mph faster than John. The time he needs to cover 12.5 miles more than John is 12.5 / 15 = 5/6 hours or 50 minutes.

Jimmy will meet John at 12:05

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

We know that speed = distance /time

==> Distance = speed * time

==> D = S*T

We will calculate the time needed foe both John and Jimmy to travel a certain distance.

Then the distance for both is the same and we will assume it is D.

Now, we know that Jogns speed = 50 mph

==> S1= 50 mph

==> The time to travel D is T1

==> D = 50 * T1............(1)

Now we will determine the equation for Jimmy.

The speed for Jimmy is S2 = 65 mph

The time needed to travel D is T2

But we know that Jimmy left 15 minutes later after John.

==> T2 = T1 - 15 minutes

15 minutes = 0.25 hour

==> T2 = T1 - 0.25

==> D = 65*(T1- 0.25)...........(2)

Now we will solve (1) and (2).

==> 50*T1 = 65*(T1 - 0.25)

==> 50T1 = 65T1 - 16.25

==> 15*T1 = 16.25

==> T1 = 16.25 / 15 =13/12

13/12 hours = 1 1/12 hours = 1 hour and 5 minutes

Then Jimmy will catch up with John after one hour and 12 minutes since John left.

John left at 11.

Then 11 + 1:05 = 12:05pm

Then Jimmy will catch up with John at 12:05 pm.