Math  /  Algebra

Question(Honors) Question 11, 7.6.47 HW Score: 50%,99\mathbf{5 0 \% , 9} 9 of 18 points Points: 0 of 1 Save
Two joggers, one averaging 8 mph and one averaging 4 mph , start from a designated initial point. The slower jogger arrives at the end of the run 40 minutes after the other jogger. Find the distance of the run.

Studdy Solution

STEP 1

1. The faster jogger averages 8 mph.
2. The slower jogger averages 4 mph.
3. The slower jogger arrives 40 minutes after the faster jogger.
4. We need to find the distance of the run.

STEP 2

1. Convert time difference into hours.
2. Define variables for time and distance.
3. Set up equations for both joggers.
4. Use the time difference to solve for the distance.

STEP 3

Convert the time difference from minutes to hours. Since there are 60 minutes in an hour, 40 minutes is:
4060=23 hours \frac{40}{60} = \frac{2}{3} \text{ hours}

STEP 4

Define variables for time and distance. Let d d be the distance of the run in miles, and t t be the time in hours it takes the faster jogger to complete the run.

STEP 5

Set up equations for both joggers based on the formula: Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}.
For the faster jogger: d=8t d = 8t
For the slower jogger, who takes 23\frac{2}{3} hours longer: d=4(t+23) d = 4(t + \frac{2}{3})

STEP 6

Use the time difference to solve for the distance. Set the two expressions for d d equal to each other:
8t=4(t+23) 8t = 4(t + \frac{2}{3})
Expand and simplify the equation:
8t=4t+83 8t = 4t + \frac{8}{3}
Subtract 4t 4t from both sides:
4t=83 4t = \frac{8}{3}
Solve for t t by dividing both sides by 4:
t=83×14=23 t = \frac{8}{3} \times \frac{1}{4} = \frac{2}{3}
Now, substitute t=23 t = \frac{2}{3} back into the equation for the faster jogger to find d d :
d=8×23=163 d = 8 \times \frac{2}{3} = \frac{16}{3}
Convert 163\frac{16}{3} miles to a decimal if needed:
d5.33 miles d \approx 5.33 \text{ miles}
The distance of the run is:
5.33 miles \boxed{5.33 \text{ miles}}

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