JEE PYQ: Motion in a Plane - Question ID 00437b0d3a73 (JEE Main 2019)

ID: 00437b0d3a73JEE Main 2019Single Correct MCQ
Ship A is sailing towards north-east with velocity v=30i+50j\mathop v\limits^ \to = 30\mathop i\limits^ \wedge + 50\mathop j\limits^ \wedge km/hr where i\mathop i\limits^ \wedge points east and j\mathop j\limits^ \wedge , north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in :

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Step-by-step Explanation

Core Formula & Concept:

In problems involving relative motion in a plane, we treat the positions and velocities of the two moving objects as vectors. The key idea is:

  • Express the position of Ship B relative to Ship A as a vector r(t)\vec{r}(t) that changes with time.
  • Express the velocity of Ship B relative to Ship A as a constant vector vrel\vec{v}_{\text{rel}}.
  • The distance between the two ships at any time tt is the magnitude of r(t)\vec{r}(t). To find the time at which this distance is minimum, we minimize r(t)2|\vec{r}(t)|^2 (which avoids dealing with square roots).

The fundamental formula is: r(t)=r0+vrelt\vec{r}(t) = \vec{r}_0 + \vec{v}_{\text{rel}} \cdot t where r0\vec{r}_0 is the initial relative position vector and vrel\vec{v}_{\text{rel}} is the relative velocity vector.

Step-by-Step Derivation:

1. Define the coordinate system and initial positions

  • Let i^\hat{i} point east and j^\hat{j} point north.
  • At t=0t=0, Ship A is at the origin: rA(0)=0\vec{r}_A(0) = 0.
  • Ship B is initially at rB(0)=80i^+150j^\vec{r}_B(0) = 80\,\hat{i} + 150\,\hat{j} km relative to Ship A.

2. Write the velocity vectors

  • Ship A’s velocity: vA=30i^+50j^\vec{v}_A = 30\,\hat{i} + 50\,\hat{j} km/hr.
  • Ship B’s velocity: vB=10i^+0j^\vec{v}_B = -10\,\hat{i} + 0\,\hat{j} km/hr (since it moves west).
  • Relative velocity of B with respect to A: vrel=vBvA=(1030)i^+(050)j^=40i^50j^  km/hr.\vec{v}_{\text{rel}} = \vec{v}_B - \vec{v}_A = (-10 - 30)\,\hat{i} + (0 - 50)\,\hat{j} = -40\,\hat{i} - 50\,\hat{j}\;\text{km/hr}.

3. Express the relative position vector at time tt

r(t)=rB(t)rA(t)=[80i^+150j^]+[40i^50j^]t.\vec{r}(t) = \vec{r}_B(t) - \vec{r}_A(t) = \bigl[80\,\hat{i} + 150\,\hat{j}\bigr] + \bigl[-40\,\hat{i} - 50\,\hat{j}\bigr]\,t.

4. Compute the squared distance D2(t)D^2(t)

D2(t)=[8040t]2+[15050t]2.D^2(t) = \bigl[80 - 40\,t\bigr]^2 + \bigl[150 - 50\,t\bigr]^2.

5. Expand and simplify

D2(t)=(8040t)2+(15050t)2=64006400t+1600t2+2250015000t+2500t2=(1600+2500)t2+(640015000)t+(6400+22500)=4100t221400t+28900.\begin{aligned} D^2(t) &= (80 - 40t)^2 + (150 - 50t)^2 \\ &= 6400 - 6400t + 1600t^2 + 22500 - 15000t + 2500t^2 \\ &= (1600 + 2500)t^2 + (-6400 - 15000)t + (6400 + 22500) \\ &= 4100\,t^2 - 21400\,t + 28900. \end{aligned}

6. Minimize D2(t)D^2(t) by calculus

To find the minimum, set the derivative with respect to tt to zero: ddtD2(t)=24100t21400=0.\frac{d}{dt}D^2(t) = 2 \cdot 4100\,t - 21400 = 0. Solving gives t=2140024100=214008200=21482=107412.6098  hours.t = \frac{21400}{2 \cdot 4100} = \frac{21400}{8200} = \frac{214}{82} = \frac{107}{41} \approx 2.6098\;\text{hours}.

7. Match with the given options

The closest option is 2.62.6 hours, which corresponds to choice C. Common Traps & Exam Tip:

1. Sign errors in relative velocity: Many students forget to subtract vA\vec{v}_A from vB\vec{v}_B or mix up the order. Always compute vrel=vBvA\vec{v}_{\text{rel}} = \vec{v}_B - \vec{v}_A. 2. Minimizing D(t)D(t) instead of D2(t)D^2(t): Differentiating the square root is messier; minimizing the squared distance is algebraically cleaner. 3. Unit consistency: Ensure all distances are in km and velocities in km/hr so the time comes out in hours. 4. Approximation errors: Carry exact fractions (e.g., 107/41107/41) until the final step to avoid rounding mistakes.

Exam Tip: When the question asks for the time of minimum distance, always set up the relative position vector and minimize its squared magnitude. This method is robust and avoids trigonometric complications.