ACSM-EP Fundamentals & Core Concepts 2 — Questions and Answers
Question 1: Using the ACSM walking equation, what is the estimated VO2 (mL/kg/min) of a client walking at 3.0 mph on a 5% grade? (1 mph = 26.8 m/min)
- 11.5
- 18.8 (Correct answer)
- 15.3
- 23.2
Correct answer: 18.8
Speed is 80.4 m/min, so VO2 = 0.1(80.4) + 1.8(80.4)(0.05) + 3.5 = 18.8 mL/kg/min.
Convert 3.0 mph to 80.4 m/min. Horizontal component: 0.1 x 80.4 = 8.04. Vertical component: 1.8 x 80.4 x 0.05 = 7.24. Add resting 3.5: 8.04 + 7.24 + 3.5 = 18.8 mL/kg/min, about 5.4 METs. The value 11.5 omits the grade term, 15.3 omits the resting component, and 23.2 incorrectly uses the running equation. The walking equation is valid for speeds of about 1.9 to 3.7 mph.
Question 2: A 70-kg client exercises at 8 METs. Approximately how many kcal/min is the client expending?
- 4.9
- 19.6
- 5.6
- 9.8 (Correct answer)
Correct answer: 9.8
kcal/min = METs x 3.5 x kg / 200 = 8 x 3.5 x 70 / 200 = 9.8 kcal/min.
Energy expenditure in kcal/min is estimated as METs x 3.5 x body mass (kg) / 200. Substituting: 8 x 3.5 x 70 = 1,960, and 1,960 / 200 = 9.8 kcal/min. This works because 1 L of O2 consumed is about 5 kcal, so VO2 of 28 mL/kg/min x 70 kg = 1.96 L/min x 5 = 9.8 kcal/min. A value of 4.9 results from dividing by 400 and 19.6 from dividing by 100.
Question 3: Using the ACSM leg cycling equation, what is the estimated VO2 (mL/kg/min) of a 70-kg client pedaling at 100 W? (1 W is about 6 kg-m/min)
- 22.4 (Correct answer)
- 18.9
- 15.4
- 9.6
Correct answer: 22.4
WR = 600 kg-m/min, so VO2 = 1.8(600)/70 + 7 = 22.4 mL/kg/min.
Convert 100 W to about 600 kg-m/min. Work component: 1.8 x 600 / 70 = 15.4. Add 7 mL/kg/min (3.5 resting plus 3.5 for unloaded cycling): 15.4 + 7 = 22.4 mL/kg/min, or 6.4 METs. Using only 3.5 for the constant gives 18.9, leaving out the 1.8 x WR/BM term gives 7, and using 100 instead of 600 for work rate gives 9.6. Always convert watts before using the equation.
Question 4: A client has an absolute VO2max of 3.0 L/min and weighs 80 kg. What is the relative VO2max in mL/kg/min?
- 26.7
- 240
- 37.5 (Correct answer)
- 375
Correct answer: 37.5
3,000 mL/min divided by 80 kg equals 37.5 mL/kg/min.
Relative VO2max = absolute VO2 in mL/min divided by body mass in kg. Converting 3.0 L/min to 3,000 mL/min gives 3,000 / 80 = 37.5 mL/kg/min, or about 10.7 METs. The value 240 multiplies instead of dividing, 26.7 divides body mass by VO2, and 375 is a decimal error. Relative values are used to compare fitness across people of different sizes, while absolute values are used for energy cost estimates.
Question 5: During maximal exercise, cardiac output is 20 L/min and the arterial-venous O2 difference is 15 mL O2 per 100 mL of blood. Using the Fick equation, what is VO2?
- 300 mL/min
- 3,000 mL/min (Correct answer)
- 35 mL/min
- 1,333 mL/min
Correct answer: 3,000 mL/min
VO2 = Q x a-vO2 difference = 20,000 mL/min x 0.15 = 3,000 mL/min.
The Fick equation states VO2 = cardiac output x (a-vO2 difference). Cardiac output is 20 L/min = 20,000 mL/min, and 15 mL per 100 mL equals 0.15. So VO2 = 20,000 x 0.15 = 3,000 mL/min (3.0 L/min). The other values come from decimal errors, adding the terms, or dividing them. Training improves VO2max mainly through increased stroke volume and cardiac output plus a wider a-vO2 difference.
Question 6: A 40-year-old client has a resting heart rate of 60 bpm and a measured maximal heart rate of 180 bpm. Using the heart rate reserve method, what is the target heart rate at 60% HRR?
- 108
- 120
- 72
- 132 (Correct answer)
Correct answer: 132
HRR = 180 - 60 = 120; 0.60 x 120 + 60 = 132 bpm.
The heart rate reserve (Karvonen) method is Target HR = [(HRmax - HRrest) x intensity] + HRrest. HRR = 180 - 60 = 120 bpm, 0.60 x 120 = 72, and adding the resting heart rate gives 132 bpm. A value of 108 is simply 60% of maximal heart rate, and 72 forgets to add resting heart rate. ACSM classifies 40 to 59% HRR (or VO2R) as moderate intensity, so this client is at the top of the moderate range.
