two people run a race. How much of a head start is needed in a second race to result in tie?

Question

John and Alex run a 100 meter race, and John wins by 9 meters.

in a second 100 meter race, how far back from the starting line does John have to be for it to result in a tie?

 

i don't know how to set this up.

Answer 1

Answer 1:

 

From the first race we can find the ratio from both distances 100 meters (d1) and 91 meters (d2).

 d1/d2 --->100/91= 1.098901

 

 

The formula of distance:  distance = speed x time   d = s . t

 

For the second race to result in a tie, it means that both runners spent the same time.

 

The formula of time is t = d/s

 

 

 

We know that in the second race Alex runs 100 meters (d2) and John (100 + X) meters (d1); being x the distance back from the starting line.

 

If both times are equal, John's time (t1) and Alex' time (t2), then:

 

t1 = t2

 

replacing the values in the formula (100+x)/d1 = 100/d2

 

 

finding a ratio between the two distances:  (100+x)/100 = d1/d2.  Replacing the value we found at the beginning +for d1/d2 (1.098901) in this equation: (100+x)/100 = 1.098901 and solving the equation we obtain x = 9.89 meters. Then John has to start 9.89 meters back from the starting line to result in a tie

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Answer 2:

 

Use the Rule of three:

 

If Alex runs 91 meters when John runs 100 meters, how many meters runs John when Alex runs 100 meters?

 

91/100 = 100/x, solving for x,  X = 109.89. this is the distance that John has to run, it means that he has to run 9.89 meters more than Alex and for that he has to star this distance back from the starting line to result in a tie

 

                                                               

Answer 2

assume John has a rate of 10m/sec

d=r*t

100=10*t

100/10=t

10 sec=t for John's time

 

Alex runs only 91 m in 10 seconds

d=r*t

91=r*10

91/10=r

9.1m/sec is Alex's rate

100=9.1*t

100/9.1=t, Alex's time in the 100 m race

10.989 sec is Alex's time in the 100 m race

d=r*t

How far would John have to run in 10.989 sec, which is Alex's time, for there to be a tie ?

(100+X)=10*(10.989)

100+X=109.89

X=109.89-100

X=9.89 m back from the starting line

Answer 3

It really helps to draw a picture:

Race 1:
          Start                                                            100 m
John:      |----------------------------------------------------|
Alex:       |-----------------------------------------| 9 m

Race 2:

                 Start                                                            100 m
John: |---x---|----------------------------------------------------|
Alex:            |----------------------------------------------------|

This is a D-I-R-T [Distance Is Rate time Time] problem. With such problems, there usually is one value that is the same or a value that is related between the two problems so we can find the answer.

For this problem:
     The times are not known (but T₁ and T₂ are different].

     The rates of each runner are assumed to not change.

     The distance run by John in Race 1 equals the distance run by Alex in Race 2.

So,

    Let x = “how far back from the starting line”

 

     [individual rates do not change, so ratio does not change]

     (John’s Rate) / (Alex’s Rate) = ( (100m)/T₁ ) / (91m/T₁) ) = ( (100m+x)/T₂ ) / (100/T₂) )

            100/91 = (100+x)/100        [cancel T₁ and T₂]

        10000 = 9100 + 91x [cross-multiply]

          900 = 91x

          9.9 m = x [rounded]

If John starts 9.9 m (rounded) back from the starting line, he and Alex will cross the finish line at the same time.