Physics Q82 – AP-EAMCET 2026

Home Physics Motion in a Straight LineQuestion #314

t = P x^{2} + Q x

is the relation between time

(t)

and distance

(x)

, where

P

and

Q

are constants, the acceleration is

Clue 1 of 3: Relationship between variables

Understand that velocity is the rate of change of position, and acceleration is the rate of change of velocity. When time is expressed as a function of position, differentiation with respect to $x$ and then using the chain rule will be necessary.

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Step 1: Find the expression for

\frac{d t}{d x}

✦ Active

The given relation between time $t$ and distance $x$ is:

t = P x^{2} + Q x

To find the velocity, we first differentiate this equation with respect to $x$ :

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

\frac{d t}{d x} = 2 P x + Q

💡 Teacher's Secret Hint

Remember that $P$ and $Q$ are constants, so their derivatives are zero when they are not multiplied by $x$ . The power rule $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ is crucial here.

Step 2: Determine the velocity (

v

)○ Expand

Velocity is defined as $v = \frac{d x}{d t}$ . We can find this by taking the reciprocal of $\frac{d t}{d x}$ :

v = \frac{d x}{d t} = \frac{1}{\frac{d t}{d x}}

v = \frac{1}{2 P x + Q}

💡 Teacher's Secret Hint

This step shows how to get $v$ when $t$ is given as a function of $x$ . This is a common technique in problems involving inverse differentiation.

Step 3: Calculate the acceleration (

a

)○ Expand

Acceleration is defined as $a = \frac{d v}{d t}$ . Since $v$ is a function of $x$ , and $x$ is a function of $t$ , we use the chain rule for differentiation:

a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t}

Since $\frac{d x}{d t} = v$ , the acceleration can also be written as:

a = v \frac{d v}{d x}

First, let's find $\frac{d v}{d x}$ from $v = (2 P x + Q)^{- 1}$ :

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

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

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

\frac{d v}{d x} = \frac{- 2 P}{(2 P x + Q)^{2}}

Now substitute $v$ and $\frac{d v}{d x}$ into the acceleration formula $a = v \frac{d v}{d x}$ :

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

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

💡 Teacher's Secret Hint

Be careful with the chain rule. Differentiating $(f (x))^{n}$ gives $n (f (x))^{n - 1} f^{'} (x)$ . Also, remember that $v = (2 P x + Q)^{- 1}$ and not just $v = (2 P x + Q)$ . A common mistake is to forget the power of $- 1$ when differentiating the velocity expression.

Step 4: Match with the options○ Expand

The calculated acceleration is $a = \frac{- 2 P}{(2 P x + Q)^{3}}$ . Comparing this with the given options, it matches option 2.

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Physics Question 82 – AP-EAMCET 2026

Clue 1 of 3: Relationship between variables

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