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Question

 

 

 

Roger can run 1 mile in 27 minutes. Jeff can do it in 24 minutes. If Jeff gives Roger 1 minute head start, how long will it take before jeff catches roger? how far will each have run?

 

 

 

 

 

it's a math problem , im trying to find the time he will take to catch roger, as well as the second part

Answer 1

Roger is the first runner and runs a distance of d₁.    Jeff is the second runner and runs a distance of d₂.

 

When Jeff catches Roger, the distance they will have run will be the same.

d₁ = d₂

 

 

If Roger runs for t minutes, Jeff runs for t-1 minutes, since Jeff gave Roger a 1 minute head start.

 

 

Distance for Roger:

d₁ = r₁*(t)

 

d₁ = (1 mi./27 minutes) * t

 

 

Distance for Jeff:

d₂ = r₂ * (t-1)  = (1mi/24 minutes) * (t-1)

 

 

 

d₁ = d₂ 

 

t/27 = (t-1)/24

 

Cross multiply:

24t = 27(t-1)

 

24t = 27t - 27

 

27 = 27t - 24t

 

27 = 3t

 

t = 9 minutes

 

 

d₁ is Roger's (and Jeff's) distance:

 

d₁ = (1 mile/27 minutes)(9 minutes) = 9/27 = 1/3 mile

 

 

Check:

d₂ is Jeff's distance

 

d2 = (1 mile/24 minutes)(9-1 minutes) = 8/24 = 1/3 mile.

 

When Jeff catches Roger, the distance they will have run will be the same.