Question

1. jacques 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? (1 mi = 5280 ft)

Explanation:

Step1: Calculate total race distance in feet

The race is 3 miles, and 1 mile = 5280 ft. So total distance $d = 3\times5280 = 15840$ ft.

Step2: Calculate distance covered in first 10 minutes

Jacques' speed in first 10 minutes is 880 ft/min. Distance $d_1 = speed\times time = 880\times10 = 8800$ ft.

Step3: Calculate distance left for second part

Distance left $d_2 = d - d_1 = 15840 - 8800 = 7040$ ft.

Step4: Calculate time for second part

Total time is 20 minutes, first part took 10 minutes, so time for second part $t_2 = 20 - 10 = 10$ minutes.

Step5: Calculate speed for second part

Speed $v_2 = \frac{d_2}{t_2} = \frac{7040}{10} = 704$ ft/min.

Answer:

704