Physics · Motion In A Straight Line · NEET
Average speed is always total distance divided by total time, not the average of the two speeds. When distances are equal, the body spends MORE time in the slower part, so the slow speed dominates the total time. Start from v_avg = total distance / total time = 2x / (x/v1 + x/v2). The x cancels, giving v_avg = 2 / (1/v1 + 1/v2) = 2v1v2/(v1+v2). The (v1+v2)/2 answer only works when the two TIMES are equal, not the distances.
The harmonic mean of two numbers is 2/(1/v1 + 1/v2). It appears because time = distance/speed, so equal distances give times proportional to 1/v1 and 1/v2. Adding times means adding reciprocals of speed, which is the definition of a harmonic mean. So 'equal distance' problems are harmonic-mean problems by construction.
Plug into v_avg = 2v1v2/(v1+v2) with v1 = v and v2 = 2v. Numerator = 2(v)(2v) = 4v^2. Denominator = v + 2v = 3v. So v_avg = 4v^2/3v = 4v/3. Notice it is closer to the slower speed v than to 2v, which confirms the slow leg took more time.
Use the arithmetic mean (v1+v2)/2 ONLY when the two TIME intervals are equal (for example, drives at v1 for 1 hour then v2 for 1 hour). Use the harmonic mean 2v1v2/(v1+v2) when the two DISTANCES are equal. Read the question: the words 'equal distance / half the distance' signal harmonic mean; 'equal time / half the time' signals arithmetic mean.
Yes, for two different positive speeds the harmonic mean is always less than or equal to the arithmetic mean (they are equal only when v1 = v2). So if you accidentally pick (v1+v2)/2 you will always get a value that is too high. This is a fast sanity check in the exam.
A vehicle travels half the distance with speed v and the remaining half distance with speed 2v. Its average speed is:
A person travelling in a straight line moves with a constant velocity v1 for a certain distance x and with a constant velocity v2 for the next equal distance. The average velocity v is given by the relation:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v_avg = 2v1v2/(v1+v2), the harmonic mean of the two speeds. It comes from total distance / total time when both distance legs are equal.
Because 2v1v2/(v1+v2) is the harmonic mean of v1 and v2. Equal distances make the times add as reciprocals of speed, which is exactly how a harmonic mean is defined.
It is always lower (or equal when v1 = v2). The slower speed occupies more time, so it pulls the average down below (v1+v2)/2.
For equal time intervals the average speed is the arithmetic mean (v1+v2)/2. For equal distances it is the harmonic mean 2v1v2/(v1+v2). Watch the wording of the question to choose correctly.
The same idea extends: v_avg = 3 / (1/v1 + 1/v2 + 1/v3) for three equal distances. But NEET almost always uses the two-leg case, so master 2v1v2/(v1+v2) first.
