A particle is rotating in a circular path and at any instant its motion can be described as theta = 5t^4/40 - t^3/3. The angular acceleration of the particle after 10 seconds is _______ rad/s^2.

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

Physics Q31 – JEE-MAIN 2026

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Home Physics Rotational MotionQuestion #429

A particle is rotating in a circular path and at any instant its motion can be described as

θ = \frac{5 t^{4}}{40} - \frac{t^{3}}{3}

. The angular acceleration of the particle after 10 seconds is _______ rad/s

^{2}

Clue 1 of 3: Kinematic Relationships

Recall that angular velocity is the rate of change of angular position, and angular acceleration is the rate of change of angular velocity.

🧠 Full Solution Path

Step 1: Determine Angular Velocity✦ Active

The angular position is given by $θ = \frac{5 t^{4}}{40} - \frac{t^{3}}{3}$ , which simplifies to $θ = \frac{t^{4}}{8} - \frac{t^{3}}{3}$ . Angular velocity $ω$ is the first derivative of angular position with respect to time $t$ :

ω = \frac{d θ}{d t} = \frac{d}{d t} (\frac{t^{4}}{8} - \frac{t^{3}}{3}) = \frac{4 t^{3}}{8} - \frac{3 t^{2}}{3} = \frac{t^{3}}{2} - t^{2}

Step 2: Determine Angular Acceleration○ Expand

Angular acceleration $α$ is the first derivative of angular velocity with respect to time $t$ :

α = \frac{d ω}{d t} = \frac{d}{d t} (\frac{t^{3}}{2} - t^{2}) = \frac{3 t^{2}}{2} - 2 t

Step 3: Calculate Angular Acceleration at t = 10 s○ Expand

Substitute $t = 10$ seconds into the expression for angular acceleration $α$ :

α (10) = \frac{3 (10)^{2}}{2} - 2 (10) = \frac{3 (100)}{2} - 20 = 150 - 20 = 130 {rad/s}^{2}

💡 Teacher's Secret Hint

Ensure correct differentiation rules are applied for each term.

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Physics Question 31 – JEE-MAIN 2026

Clue 1 of 3: Kinematic Relationships

🧠 Full Solution Path