JEE PYQ: Motion in a Straight Line - Question ID 064200b00d3d (JEE Main 2021)

Core Formula & Concept:

In kinematics, the instantaneous velocity $V(t)$ of a particle moving along a straight line is the time derivative of its displacement $s(t)$: $V(t) = \frac{ds}{dt}$ Conversely, the displacement $s(t)$ can be obtained by integrating the velocity function with respect to time: $s(t) = \int V(t) \, dt + C$ where $C$ is the constant of integration, determined by initial conditions.

The distance travelled between two time instants $t_1$ and $t_2$ is the integral of the magnitude of velocity over that interval: $\text{Distance} = \int_{t_1}^{t_2} |V(t)| \, dt$ However, if the velocity $V(t)$ does not change sign in the interval $[t_1, t_2]$, then the distance travelled is simply the absolute difference of the displacement at $t_2$ and $t_1$: $\text{Distance} = |s(t_2) - s(t_1)|$

In this problem, the velocity is given as: $V(t) = \alpha t + \beta t^2$ We are to find the distance travelled between $t = 1\,s$ and $t = 2\,s$. We must first check whether $V(t)$ changes sign in this interval. If it does not, we can directly integrate $V(t)$ to find displacement and use it to compute distance.

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Step-by-Step Derivation:

Step 1: Integrate the velocity function to find displacement

We are given: $V(t) = \alpha t + \beta t^2$ Integrate $V(t)$ with respect to $t$ to find displacement $s(t)$: $s(t) = \int V(t) \, dt = \int (\alpha t + \beta t^2) \, dt = \frac{\alpha}{2} t^2 + \frac{\beta}{3} t^3 + C$ where $C$ is the integration constant. Since we are interested in the difference in displacement between two times, the constant $C$ cancels out and can be ignored.

Step 2: Evaluate displacement at $t = 1\,s$ and $t = 2\,s$

Compute $s(1)$ and $s(2)$: $s(1) = \frac{\alpha}{2} (1)^2 + \frac{\beta}{3} (1)^3 = \frac{\alpha}{2} + \frac{\beta}{3}$ $s(2) = \frac{\alpha}{2} (2)^2 + \frac{\beta}{3} (2)^3 = \frac{\alpha}{2} \cdot 4 + \frac{\beta}{3} \cdot 8 = 2\alpha + \frac{8\beta}{3}$

Step 3: Compute the displacement difference

The displacement from $t = 1$ to $t = 2$ is: $\Delta s = s(2) - s(1) = \left(2\alpha + \frac{8\beta}{3}\right) - \left(\frac{\alpha}{2} + \frac{\beta}{3}\right)$ Simplify: $\Delta s = 2\alpha - \frac{\alpha}{2} + \frac{8\beta}{3} - \frac{\beta}{3} = \frac{4\alpha - \alpha}{2} + \frac{8\beta - \beta}{3} = \frac{3\alpha}{2} + \frac{7\beta}{3}$

Step 4: Verify that velocity does not change sign in $[1, 2]$

We must ensure that $V(t) = \alpha t + \beta t^2$ does not cross zero in $t \in [1, 2]$. Since $t \geq 1$, $t^2 \geq t$, and if $\alpha$ and $\beta$ are positive (typical in such problems unless stated otherwise), $V(t)$ is positive throughout. Even if $\alpha$ or $\beta$ are negative, unless specified, we assume the velocity does not change sign in the interval. Thus, the distance travelled is equal to the magnitude of the displacement difference. Therefore: $\text{Distance} = \left| \Delta s \right| = \frac{3}{2}\alpha + \frac{7}{3}\beta$

Step 5: Match with given options

Comparing with the options: - A: $3\alpha + 7\beta$ - B: $\frac{3}{2}\alpha + \frac{7}{3}\beta$ - C: $\frac{\alpha}{2} + \frac{\beta}{3}$ - D: $\frac{3}{2}\alpha + \frac{7}{2}\beta$ Our result matches Option B. ---

Common Traps & Exam Tip:

Trap 1: Confusing displacement with distance. Many students forget to check whether the velocity changes sign. If $V(t)$ crosses zero, the distance is not simply $|s(2) - s(1)|$, but the sum of absolute values of integrals over sub-intervals where $V(t)$ is positive or negative. In this case, since no information suggests a sign change, we proceed under the assumption that $V(t)$ does not change sign.

Trap 2: Incorrect integration. Students often make errors in integrating $t^2$, mistakenly writing $\frac{t^3}{2}$ instead of $\frac{t^3}{3}$. Always double-check integration rules.

Trap 3: Ignoring the constant of integration. While $C$ cancels out in displacement differences, forgetting it during integration can lead to confusion. Remember: for definite integrals, constants cancel.

Exam Tip: When given velocity as a function of time, always:

1. Integrate to find displacement.
2. Check if velocity changes sign in the interval.
3. If no sign change, distance = $|s(t_2) - s(t_1)|$.
4. Simplify carefully and match with options.

In this question, following these steps leads you confidently to Option B.