JEE Physics 2024 Question with Solution

Given: $t = \alpha x^2 + \beta x$

Find: Relation between acceleration $a$ and velocity $v$.

Differentiate the given relation with respect to $x$:

$$\frac{dt}{dx} = 2\alpha x + \beta$$

Since

$$v = \frac{dx}{dt}$$

we get

$$v = \frac{1}{\frac{dt}{dx}} = \frac{1}{2\alpha x + \beta}$$

Now acceleration is

$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$$

Differentiate $v$ with respect to $x$:

$$\frac{dv}{dx} = -\frac{2\alpha}{(2\alpha x + \beta)^2}$$

Therefore,

$$a = v \cdot \frac{dv}{dx} = \left(\frac{1}{2\alpha x + \beta}\right)\left(-\frac{2\alpha}{(2\alpha x + \beta)^2}\right) = -\frac{2\alpha}{(2\alpha x + \beta)^3}$$

Using

$$v = \frac{1}{2\alpha x + \beta}$$

we obtain

$$a = -2\alpha v^3$$

Therefore, the correct option is A.

Given: $t = \alpha x^2 + \beta x$

Find: Relation between $a$ and $v$.

Differentiate with respect to $x$:

$$\frac{dt}{dx} = 2\alpha x + \beta$$

Since $\frac{dt}{dx} = \frac{1}{v}$, we have

$$\frac{1}{v} = 2\alpha x + \beta$$

Differentiate this relation with respect to time:

$$-\frac{1}{v^2}\frac{dv}{dt} = 2\alpha \frac{dx}{dt}$$

Using $\frac{dx}{dt} = v$ and $\frac{dv}{dt} = a$,

$$-\frac{a}{v^2} = 2\alpha v$$

Hence,

$$a = -2\alpha v^3$$

This method works because expressing $\frac{dt}{dx}$ as $\frac{1}{v}$ avoids differentiating a reciprocal function explicitly. Therefore, the correct option is A.