Consider a particle moving along a straight line, whose position as a function of time is given by s(t)=αt²-βt+γ, where α=1 ms⁻², β=6 ms⁻¹ and γ=5 m. The average speed of the particle, in ms⁻¹, from t=0 to t=6 s is :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q6 – NEET-UG 2026

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Home Physics Motion in a Straight LineQuestion #333

Consider a particle moving along a straight line, whose position as a function of time is given by

s (t) = α t^{2} - β t + γ

, where

α = 1 {ms}^{- 2}

β = 6 {ms}^{- 1}

and

γ = 5 m

. The average speed of the particle, in

{ms}^{- 1}

, from

t = 0

t = 6 s

is :

Clue 1 of 3: Definition of Average Speed

Average speed is defined as the total distance traveled divided by the total time taken.

🥷

Ninja Strategy•Distinguish Average Speed from Average Velocity

Recognize that average speed requires calculating total distance, which involves considering changes in direction, while average velocity only depends on net displacement. Since the particle returns to its initial position, its average velocity is zero, making option (4) a common distractor for average speed.

🧠 Full Solution Path

Step 1: Determine the Velocity Function and Turning Points✦ Active

The position of the particle is given by $s (t) = α t^{2} - β t + γ$ . Substitute the given values $α = 1 {ms}^{- 2}$ , $β = 6 {ms}^{- 1}$ , and $γ = 5 m$ into the equation:

s (t) = 1 t^{2} - 6 t + 5

To find the velocity function, differentiate $s (t)$ with respect to $t$ :

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

The particle changes direction when its velocity is zero. Set $v (t) = 0$ to find the turning point:

2 t - 6 = 0 ⟹ 2 t = 6 ⟹ t = 3 s

Since $t = 3 s$ is within the interval $[0, 6 s]$ , the particle changes direction at this time.

💡 Teacher's Secret Hint

Remember that average speed requires total distance, not just net displacement. A change in direction means you must sum the magnitudes of displacements in each segment.

Step 2: Calculate Total Distance Traveled○ Expand

Calculate the position of the particle at $t = 0 s$ , $t = 3 s$ , and $t = 6 s$ :

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

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

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

Now, calculate the distance traveled in each segment:

Distance from $t = 0 s$ to $t = 3 s$ :

d_{1} = | s (3) - s (0) | = | - 4 - 5 | = | - 9 | = 9 m

Distance from $t = 3 s$ to $t = 6 s$ :

d_{2} = | s (6) - s (3) | = | 5 - (- 4) | = | 9 | = 9 m

The total distance traveled is the sum of these distances:

D_{t o t a l} = d_{1} + d_{2} = 9 m + 9 m = 18 m

💡 Teacher's Secret Hint

Always take the absolute value of displacement for each segment to ensure you are calculating distance, not just change in position.

Step 3: Calculate Average Speed○ Expand

The total time taken is $Δ t = 6 s - 0 s = 6 s$ . Average speed is calculated as total distance divided by total time:

Average Speed = \frac{D_{t o t a l}}{Δ t} = \frac{18 m}{6 s} = 3 {ms}^{- 1}

💡 Teacher's Secret Hint

Be careful not to confuse average speed with average velocity. Average velocity is total displacement divided by total time, which would be zero in this case since $s (6) = s (0)$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 6 – NEET-UG 2026

Clue 1 of 3: Definition of Average Speed

🧠 Full Solution Path