Brad and Curtis went walking. Brad began walking 20 seconds earlier than Curtis. Brad walks at a rate of 5 feet per second. Curtis walks at a rate of 6 feet

Question

Two boys, Brad and Curtis, went for a walk. Brad began walking 20 seconds earlier than Curtis. Brad walked at a speed of 5 feet per second. Curtis walked at a speed of 6 feet per second. For how many seconds had Brad been walking at the moment when the two boys had walked exactly the same distance?

 

Brad and Curtis went walking.  Brad began walking 20 seconds earlier than Curtis.  Brad walks at a rate of 5 feet per second.  Curtis walks at a rate of 6 feet per second.  How many seconds had Brad walked, when himself and Curtis had walked to same distance? 

 

 

Answer 1

With work problem, there are few standard types (you should get familiar with each type).  This one is called “distance – rate – time.”  The acronym I like is “D = R*T”  (you know, dirt).  That fits because “is” translates to ‘=’ in math.

Usually, there are two boys, two trains,… two somethings.  They each follow their own “dirt” so that makes two equations.  The problem also tells us how they interact – either one of the variables (D, R, or T) are the same or else the two of them travel a total distance that is given.

What do we have in this problem? (let B=Brad’s distance,  C=Curtis’s distance, T = Brad’s start time)
The distances are the same:

     B = C  

And  putting in the rates and times:
    (5 ft/sec) * T sec  =  (6 ft/sec)*(T-20) sec      [Curtis’s time is 20 sec less than Brad’s]
     5T = 6T – 120
      120 seconds  =  T            the time Brad had been walking when they met

Checking:
     Is      5*120 = 6*(100)   ?    yes