Physics · Motion In A Straight Line · NEET
Yes, exactly. Average speed = total distance / total time. If each speed lasts time t: total distance = v1·t + v2·t = (v1 + v2)t, and total time = t + t = 2t. So average speed = (v1 + v2)t / 2t = (v1 + v2)/2. The time t cancels no matter what its value is, so you never even need to know t. This simple mean works ONLY when the two time intervals are equal.
They weight the slow part differently. In equal TIME, the fast speed and slow speed each act for the same duration, so both count equally — you just average them. In equal DISTANCE, the slow speed takes MORE time to cover its half, so it dominates the total time, pulling the average down. That is why equal distance needs the harmonic mean 2v1v2/(v1+v2), which is always smaller than or equal to the simple mean (v1+v2)/2. Rule: same time = arithmetic mean; same distance = harmonic mean.
Same idea: just add all the speeds and divide by how many there are. For three equal times at v1, v2, v3: average speed = (v1 + v2 + v3)/3. The equal times cancel out again. So for n equal time intervals, average speed = (v1 + v2 + ... + vn)/n. This is why the equal-time case is the 'easy' one in NEET — no harmonic mean, no fractions.
Only if the motion stays in one straight line without turning back. Average speed uses total distance (always positive). Average velocity uses displacement, which can cancel if direction reverses. If the object moves forward at v1 and forward at v2 for equal times, average speed = average velocity = (v1+v2)/2. But if it reverses direction, the distances still add for speed while displacements subtract for velocity, so the two answers differ. Always check the direction before reusing the formula.
A toy car with charge q moves on a frictionless horizontal plane under a uniform electric field E. Due to force qE its velocity increases from 0 to 6 m/s in one second. At that instant the field is reversed and the car moves for two more seconds. The average velocity and average speed of the car between 0 and 3 seconds are respectively:
A particle moves along a straight line with position s(t) = (alpha)t^2 - (beta)t + gamma, where alpha = 1 m/s^2, beta = 6 m/s, gamma = 5 m. The average speed of the particle from t = 0 to t = 6 s is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Average speed = (v1 + v2)/2, the simple arithmetic mean of the two speeds. It works because the equal time t cancels from total distance (v1+v2)t and total time 2t.
Do not use it when the two parts cover equal DISTANCE instead of equal time. For equal distance use the harmonic mean 2v1v2/(v1+v2). Also do not treat it as average velocity if the direction reverses.
Yes. The arithmetic mean (v1+v2)/2 is always greater than or equal to the harmonic mean 2v1v2/(v1+v2). They are equal only when v1 = v2. So equal-time average speed is never smaller than equal-distance average speed.
Just take the ordinary average of all the speeds: (v1 + v2 + ... + vn)/n for n equal time intervals. Every equal time cancels, so no weighting is needed.
Because one word ('time' vs 'distance') changes the whole formula. NEET checks whether you read the split correctly rather than doing hard math. Getting the mean type right earns an easy mark.
