Consider a cardiac monitor whose time-base control is set so that the ECG trace moves horizontally at 25 mm/s. If the patient’s heart rate is 75 beats/min, how far apart would the peaks of the ECG trace be?
A. 15 mm
B. 18 mm
C. 20 mm
D. 25 mm
The correct answer and explanation is :
To determine the distance between the peaks of the ECG trace, we need to follow these steps:
Step 1: Calculate the Period of the Heartbeat
The period of a heartbeat (( T )) is the time between successive beats. It is the reciprocal of the heart rate:
[
T = \frac{60 \text{ s/min}}{\text{Heart Rate (beats/min)}}
]
Given that the heart rate is 75 beats per minute:
[
T = \frac{60}{75} = 0.8 \text{ seconds per beat}
]
Step 2: Determine the Distance Traveled on the ECG Paper
The ECG trace moves at a speed of 25 mm per second. The total distance moved in one beat (or one period) is:
[
\text{Distance} = \text{Speed} \times \text{Time}
]
[
= (25 \text{ mm/s}) \times (0.8 \text{ s})
]
[
= 20 \text{ mm}
]
Step 3: Identify the Correct Answer
The peaks of the ECG trace, which correspond to successive R-waves in a normal ECG, will be 20 mm apart.
Thus, the correct answer is:
[
\mathbf{C. \ 20 \ mm}
]
Explanation
The ECG paper moves at a standard speed of 25 mm per second, meaning that for every second, the trace advances 25 mm. The heart beats at 75 beats per minute, meaning a single beat takes 0.8 seconds. During this 0.8-second interval, the paper moves 20 mm, which represents the distance between successive R-wave peaks in the ECG.
This calculation is useful for interpreting ECGs, as the distance between the R-waves (RR interval) helps in determining heart rate and diagnosing abnormalities such as tachycardia or bradycardia.