Physics Q15 – NEET-UG 2025

Home Physics Motion in a Straight LineQuestion #327

In some appropriate units, time

(t)

and position

(x)

relation of a moving particle is given by

t = x^{2} + x

. The acceleration of the particle is

Clue 1 of 3: Kinematic Definitions

Recall that velocity is the first derivative of position with respect to time, and acceleration is the first derivative of velocity with respect to time (or the second derivative of position).

🧠 Full Solution Path

● Video Walkthrough

Step 1: Find Velocity (

v

)✦ Active

Given the relation between time $(t)$ and position $(x)$ as $t = x^{2} + x$ . To find the velocity $v = \frac{d x}{d t}$ , differentiate the given equation with respect to $t$ :

\frac{d}{d t} (t) = \frac{d}{d t} (x^{2} + x)

1 = 2 x \frac{d x}{d t} + \frac{d x}{d t}

1 = (2 x + 1) \frac{d x}{d t}

v = \frac{d x}{d t} = \frac{1}{2 x + 1} = (2 x + 1)^{- 1}

💡 Teacher's Secret Hint

Remember to apply the chain rule when differentiating $x$ with respect to $t$ .

Step 2: Find Acceleration (

a

)○ Expand

Now, to find the acceleration $a = \frac{d v}{d t}$ , differentiate the velocity expression $v = (2 x + 1)^{- 1}$ with respect to $t$ . We use the chain rule $a = \frac{d v}{d x} \cdot \frac{d x}{d t}$ :

\frac{d v}{d x} = \frac{d}{d x} ((2 x + 1)^{- 1}) = - 1 (2 x + 1)^{- 2} \cdot (2) = - 2 (2 x + 1)^{- 2}

a = \frac{d v}{d t} = (- 2 (2 x + 1)^{- 2}) \cdot (\frac{1}{2 x + 1})

a = - 2 (2 x + 1)^{- 2} (2 x + 1)^{- 1}

a = - 2 (2 x + 1)^{- 3}

a = - \frac{2}{(2 x + 1)^{3}}

Comparing this result with the given options, option (2) matches.

💡 Teacher's Secret Hint

Be careful with the negative exponent and the chain rule when differentiating $(2 x + 1)^{- 1}$ .

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

Physics Question 15 – NEET-UG 2025

Clue 1 of 3: Kinematic Definitions

🧠 Full Solution Path