Question

Jack ran at 880 ft./min. For the first 10 min. Of the 3 mi. Race before slowing his pace to finish the race with a total time of 20 min. What was his pace (in feet per minute) for the second part of the race?

Answer 1

Answer:

Jack ran 704 ft/min during the second half of the race.

Step-by-step explanation:

Here the distance traveled was 3 miles, and the total elapsed time was 20 minutes.  Recall that distance = rate * time, or rate = distance / time.

Find the pace (speed, rate) for the second part of the race.

Here, the total distance was 3 miles and the elapsed time 20 minutes.  Jack's average speed was thus 3/20 miles/minute.  Converting this to feet per minute, we get:

3 mi          5280 ft

------------ * ------------- = 792 ft/min    (Jack's average speed)

20 min        1 mi

Recall that the formula for finding the average of x and y is (x + y) / 2.

(1/2)(880 ft/min  +  p) = 792 ft/min; find p, which represents Jack's pace for the second part of the race.

If (1/2)(880 ft/min  +  p) = 792 ft/min, and we mult. both sides by 2, we get:

880 ft/min + p = 1584 ft/min.

To solve this for p, subtract 880 ft/min from both sides, obtaining:

p = 704 ft/min

Jack ran 704 ft/min during the second half of the race.