JEE PYQ: Motion in a Straight Line - Question ID 787e7d3dce51 (JEE Main 2022)

Core Formula & Concept:

In kinematics, when we are given a relationship between position $x$ and time $t$, we often need to find the instantaneous velocity $v = \frac{dx}{dt}$. The key concept here is differentiation with respect to time. Specifically, if $t$ is expressed as a function of $x$, we can use the chain rule of differentiation to find $\frac{dx}{dt}$ as the reciprocal of $\frac{dt}{dx}$:

$\frac{dx}{dt} = \frac{1}{\frac{dt}{dx}}.$

This approach is particularly useful when the given equation is not explicitly solved for $x$ in terms of $t$, but rather $t$ is given in terms of $x$.

Step-by-Step Derivation:

Step 1: Write down the given relation.

$t = \sqrt{x} + 4.$

Step 2: Differentiate both sides with respect to $x$.

We want to find $\frac{dt}{dx}$. Differentiating the right-hand side:

$\frac{dt}{dx} = \frac{d}{dx} \left( \sqrt{x} + 4 \right) = \frac{d}{dx} \left( x^{1/2} \right) + \frac{d}{dx} (4).$

Using the power rule and the fact that the derivative of a constant is zero:

$\frac{dt}{dx} = \frac{1}{2} x^{-1/2} + 0 = \frac{1}{2 \sqrt{x}}.$

Step 3: Express $\frac{dx}{dt}$ using the reciprocal.

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

$\frac{dx}{dt} = \frac{1}{\frac{1}{2 \sqrt{x}}} = 2 \sqrt{x}.$

Step 4: Find the value of $x$ when $t = 4$.

Substitute $t = 4$ into the original equation:

$4 = \sqrt{x} + 4.$

Subtract 4 from both sides:

$0 = \sqrt{x}.$

Square both sides:

$x = 0.$

Step 5: Evaluate $\frac{dx}{dt}$ at $t = 4$ (i.e., at $x = 0$).

Substitute $x = 0$ into the expression for $\frac{dx}{dt}$:

$\left( \frac{dx}{dt} \right)_{t=4} = 2 \sqrt{0} = 0.$

Conclusion:

The value of $\left( \frac{dx}{dt} \right)_{t=4}$ is $0$, which corresponds to option B.

Common Traps & Exam Tip:

Trap 1: Misapplying the chain rule. Some students mistakenly differentiate $t$ with respect to $t$ instead of $x$, leading to incorrect expressions like $\frac{dt}{dt} = 1$. Always ensure you differentiate with respect to the correct variable.

Trap 2: Forgetting to find $x$ at $t = 4$. Directly substituting $t = 4$ into $\frac{dx}{dt} = 2 \sqrt{x}$ without first finding $x$ at $t = 4$ can lead to confusion. Always solve for $x$ first when $t$ is given.

Trap 3: Ignoring the domain of $\sqrt{x}$. The square root function $\sqrt{x}$ is only defined for $x \geq 0$. At $t = 4$, we find $x = 0$, which is valid but leads to a zero velocity. Students might overlook this and assume a non-zero value.

Exam Tip: When given $t$ as a function of $x$, always differentiate $t$ with respect to $x$ first, then take the reciprocal to find $\frac{dx}{dt}$. This method is more straightforward and avoids errors in implicit differentiation.