A tortoise is moving at a rate of 2 feet per minute, while a hare is running at a rate of 8 feet per minute.

Question

A tortoise is moving at a rate of 2 feet per minute, while a hare is running at a rate of 8 feet per minute. If the tortoise had a head start of 180 feet, how long will it take for the tortoise and hare to meet?
Which variables would you use for the equations of this scenario? Check all that apply.

Answer 1

I'll describe, but not solve, the problem.

 

This is a D-I-R-T [Distance-Is-Rate-times-Time] problem] -- remember that formula.

 

180 feet and the distance each runs (assign a variable) are the important distances.

   Tortise Start               Rabbit Start                                         Catch Up

        | - - - - - -- - - - - - - - | - - - - - - - - - - - - - - - - - - - - - - - - |

                 180 ft.                                  d ft

 

 [note: you could assign d to be the distance the rabbit runs and (d+180) as the distance for the tortise; or

           you could assign  d as the distance the distance the tortise runs and (d-180) as the distance for the rabbit, but

           remember which is d when you want to find the other distance.

 

The question asks "How long until they meet?"  This is the time.  You must determine whether "how long" starts from the time the rabbit starts or from the time the tortoise started and the rabbit waited.  Since the problem gives the rates in terms of "is running," it makes sense to have t=0 be the time the rabbit starts.  That means that d=rabbit's distance makes sense, too (and tortoise distance is d+180).  PLZ note that this determines the two D-I-R-T formulas that are equal:

    Rabbit's distance = Tortise's distance + 180 ft

 

Now, as is often the case with D-I-R-T problems, we must find one variable (d in this problem) before we may find the other variable (t).

 

Special note:  Watch out for any difference in units!  Some problems may trick you, for example, if the tortoise runs at 2 feet/minute and the rabbit runs at 8 feet/second.

 

So, when you find the value of one variable, you may use either D-I-R-T equation to find the other variable and answer the problem.

 

A very important practice:  Use your variable values to check that the D-I-R-T formulas are, in fact, equal.

 

Answer 2

d = r · t

d = 2t (tortoise)

d + 180 = 8t (rabbit)

 

d = 2t

d = 8t - 180

 

2t = 8t - 180

 

Can you solve for t and answer?

 

 

Answer 3

Let x = number of minutes run by the tortoise since the hare started running

 

Distance = (Rate)(Time)

 

Distance (in feet) of tortoise since the hare started running = 180 + 2x

Distance (in feet) run by the hare in x minutes = 8x

 

The hare catches up to the tortoise when 8x = 180 + 2x

 

                                                            6x = 180

 

                                                             x = 30 minutes

 

The hare catches up to the tortoise in 30 minutes ( Half an hour).