Kinematics NEET 2026: Average Speed with Direction Reversal

What is the answer to the Kinematics NEET 2026 average-speed question?

Option C, 3 metres per second, is correct: in this Kinematics NEET 2026 question, the particle returns to its starting position but travels 18 metres. The key is finding where it reverses direction, rather than treating displacement as distance.

A particle moves along a straight line with the position law: $s (t) = α t^{2} - β t + γ$

α = 1 m s^{- 2}, β = 6 m s^{- 1}, γ = 5 m

Find its average speed during the first 6 seconds, starting at zero time. The choices are:

12 m s^{- 1}

6 m s^{- 1}

3 m s^{- 1}

0 m s^{- 1}

The question bank classifies this as medium, with an expected solving time of 90 seconds. That is the question bank’s practice target, not an official examination time limit.

What do the starting and finishing positions tell us?

The particle starts and finishes at the same coordinate, so its displacement is zero. That establishes zero average velocity, but not zero average speed. Following the official solution, check the endpoints first, then examine the motion between them.

Substituting the constants gives the numerical position law, with time in seconds and position in metres: $s (t) = t^{2} - 6 t + 5$

At the start:

s (0) = 0^{2} - 6 (0) + 5 = 5 m

At the finish:

s (6) = 6^{2} - 6 (6) + 5 = 36 - 36 + 5 = 5 m

The displacement is:

Displacement = s (6) - s (0) = 5 - 5 = 0 m

Equal endpoints do not establish zero distance travelled. The particle could have moved away and returned, so we must check its direction during the interval.

How do we find and verify the direction reversal?

The particle reverses direction at 3 seconds, when its velocity changes from negative to positive. Finding zero velocity alone is not enough: the sign change confirms that we must split the journey into separate distance calculations.

Velocity is the time derivative of position. Differentiate each term:

v (t) = \frac{d s}{d t} = \frac{d}{d t} (t^{2}) - \frac{d}{d t} (6 t) + \frac{d}{d t} (5)

$v (t) = 2 t - 6 + 0 = 2 t - 6$ Set velocity equal to zero: $2 t - 6 = 0$ $2 t = 6$ $t = 3 s$

This time lies inside the requested interval from zero to 6 seconds. Check the velocity signs on both sides:

0 \leq t < 3 : v (t) = 2 t - 6 < 0

3 < t \leq 6 : v (t) = 2 t - 6 > 0

Before 3 seconds, the coordinate decreases; after 3 seconds, it increases. The journey therefore has two one-direction segments, meeting at this turning-point position:

s (3) = 3^{2} - 6 (3) + 5 = 9 - 18 + 5 = - 4 m

A negative position is a coordinate relative to the origin, not a negative distance travelled. The particle reaches a point 4 metres on the negative side of the origin before returning.

How do we calculate total distance and average speed?

The two segments each cover 9 metres, giving a total distance of 18 metres in 6 seconds. On either side of the reversal, velocity keeps the same sign. Each segment’s distance therefore equals the absolute change in position.

From zero to 3 seconds:

d_{1} = | 5 - (- 4) | = 9 m

From 3 to 6 seconds:

d_{2} = | - 4 - 5 | = 9 m

Add the distances, not the signed changes in position:

Total distance = 9 + 9 = 18 m

The elapsed time is: $Δ t = 6 - 0 = 6 s$

Therefore:

Average speed = \frac{Total distance}{Elapsed time} = \frac{18}{6} = 3 m s^{- 1}

The final answer is option C. The outward and return distances add because both are lengths travelled; they do not cancel as displacements do.

For another interval, check endpoints, find any velocity sign changes within the interval, add segment distances, then divide by total time.

Why does option D give average velocity rather than average speed?

Option D comes from dividing displacement by elapsed time. That correctly calculates average velocity, but it answers the wrong question here. The error is the numerator, not the arithmetic.

The incorrect calculation for average speed is:

\frac{s (6) - s (0)}{6} = \frac{5 - 5}{6} = 0 m s^{- 1}

For this motion, the results are:

Displacement: zero metres, because the endpoint coordinates match.
Average velocity: zero metres per second, because net displacement is zero.
Distance: 18 metres, counting both parts of the journey.
Average speed: 3 metres per second, using that total distance.

Taking the absolute value of the final displacement does not repair the error:

| s (6) - s (0) | = | 5 - 5 | = 0 m

Apply absolute values to the position changes of separate one-direction segments before adding. The particle is momentarily at rest at 3 seconds, not stationary throughout the interval. Its velocity is nonzero before and after that instant.

How does the answer change when the interval includes the reversal?

Without a reversal inside the interval, one absolute endpoint difference gives the distance. With a reversal, split the motion there and add the segment distances. These are original related practice questions, not additional verified NEET PYQs.

What is the average speed from zero to 2 seconds?

The answer is 4 metres per second because velocity remains negative throughout this interval. Original related practice, Question 1: For the same position law, find average speed from zero to 2 seconds.

v (t) = 2 t - 6 < 0 (0 \leq t \leq 2)

s (0) = 5 m, s (2) = 4 - 12 + 5 = - 3 m

Distance = | - 3 - 5 | = 8 m

Δ t = 2 - 0 = 2 s

Average speed = \frac{8}{2} = 4 m s^{- 1}

What are the average speed and average velocity from 2 to 4 seconds?

Average speed is 1 metre per second, while average velocity is zero metres per second: this interval contains the reversal at 3 seconds. Original related practice, Question 2: Find both averages for the same particle from 2 to 4 seconds.

s (2) = - 3 m, s (3) = - 4 m

s (4) = 16 - 24 + 5 = - 3 m

Distance = | - 4 - (- 3) | + | - 3 - (- 4) | = 1 + 1 = 2 m

Δ t = 4 - 2 = 2 s

Average speed = \frac{2}{2} = 1 m s^{- 1}

Average velocity = \frac{- 3 - (- 3)}{4 - 2} = 0 m s^{- 1}

The first interval needs no split; the second must be split at the reversal. Before dividing, write the elapsed time, not the final clock reading, in the denominator.

Next step: photograph a doubt on NEET JEEnius AI and photograph any question you are stuck on and get a step-by-step solution across Physics, Chemistry and Biology (20 free a month).

Related on NEET JEEnius AI: How to Study Kinetic Theory NEET: 7 Steps for MCQs.

Frequently asked questions

What is the average speed for s(t) = t^2 - 6t + 5 in the first 6 seconds?

With position in metres and time in seconds, the average speed is 3 m/s, so option C is correct. The particle moves from 5 m to -4 m and then returns to 5 m, covering 18 m in 6 s.

How do I find when a particle reverses direction?

Differentiate the position function to find velocity, then locate its zeros inside the time interval. Check that velocity changes sign across each zero; zero velocity alone does not establish a reversal. For s(t) = t^2 - 6t + 5, velocity is 2t - 6 and changes from negative to positive at 3 s.

Can average velocity be zero when average speed is not zero?

Yes: returning to the starting position gives zero displacement and therefore zero average velocity, even if the particle has travelled a nonzero distance. For s(t) = t^2 - 6t + 5 over the first 6 s, displacement is zero but distance is 18 m. Average velocity is therefore zero, while average speed is 3 m/s.

When can I use the absolute change in position to find distance?

Use the absolute endpoint difference when there is no direction reversal inside the interval. If the particle reverses direction, split the interval at each reversal and add the absolute position changes of the separate one-direction segments. Taking the absolute value of net displacement can miss the distance covered before returning.