Jay left Durham, and 3 hrs later, Mike left going 50 mph faster to catch up. After another 2 hrs, Mike caught up. Find Jay's average speed.

Question

please answer this with a solution and equation. thank you!

Answer 1

Emma:

 

Assume that Mike also left from Durham!

 

Let d be the distance (miles) Jay and Mike traveled. Note that Jay has been traveling for 5 hours, before Mike caught up with him. Let U mph be Jay's speed. Then d = 5U.

 

Mike's speed was (U + 50) mph.  We are told that Mike left 3 hrs later and after another 2 hrs Mike caught up. So Mike had traveled the same distance d in 2 hours at a speed of U + 50.  So d = 2(U+50).  Equating the two expressions for d, we get

 

d = 5U = 2(U+50)

 

This gives U = 100/3 = 33 + (1/3)

 

Back check.  Jay has been traveling for 5 hours to cover a distance of 5(100/3) = 500/3

                  Mike's speed is 100/3 + 50 = 250/3.  In only 2  hours he covered 2(250/3) = 500/3  to catch up!

 

Hurray!

 

Dattaprabhakar (Dr.G.)

 

Answer 2

Let v be the Jay's average speed. Then (v + 50)mph is the Mike's average speed. When they met Jay was traveling 5 hours, while Mike has spent just 2 hours. To meet at the same point means to travel the same distance, because the starting is also the same. Thus, we have

 

                                            5v = 2(v+ 50)                                             (1)

or

                                               3v  = 100 mph

Hence

                                     v = 100/3 = 33.33 mph

 

Plug these numbers into equation (1) to make sure that both sides are equal.