JEE PYQ: Motion in a Straight Line - Question ID 3bd124967b11 (JEE Main 2022)

Core Formula & Concept:

When a body moves vertically under gravity, its motion is uniformly accelerated (or decelerated). The key equations are:

• Displacement as a function of time: $s(t) = ut - \tfrac{1}{2}gt^2$ where $u$ is the initial upward velocity, $g = 10\ \text{ms}^{-2}$, and $t$ is the time elapsed since projection.
• Velocity as a function of time: $v(t) = u - gt$

In this problem, two balls are projected upward with the same initial speed $u = 50\ \text{ms}^{-1}$, but at different times. The first ball is launched at $t = 0$, and the second at $t = 2\ \text{s}$. We need to find the instant $t$ (measured from $t = 0$) when the two balls meet.

Step-by-Step Derivation:

Step 1: Define the time variables

Let $t$ be the time measured from the start ($t = 0$). The first ball has been in motion for $t$ seconds, while the second ball has been in motion for $t - 2$ seconds (since it was launched at $t = 2$).

Step 2: Write the displacement equations

Displacement of the first ball at time $t$: $s_1(t) = 50t - \tfrac{1}{2} \cdot 10 \cdot t^2 = 50t - 5t^2$

Displacement of the second ball at time $t$ (it has been moving for $t - 2$ seconds): $s_2(t) = 50(t - 2) - \tfrac{1}{2} \cdot 10 \cdot (t - 2)^2 = 50(t - 2) - 5(t - 2)^2$

Step 3: Set the displacements equal at the meeting point

At the instant they meet, $s_1(t) = s_2(t)$: $50t - 5t^2 = 50(t - 2) - 5(t - 2)^2$

Step 4: Expand and simplify

Expand the right side: $50(t - 2) - 5(t^2 - 4t + 4) = 50t - 100 - 5t^2 + 20t - 20$ $= (50t + 20t) - 100 - 20 - 5t^2$ $= 70t - 120 - 5t^2$

Now the equation becomes: $50t - 5t^2 = 70t - 120 - 5t^2$

Cancel $-5t^2$ from both sides: $50t = 70t - 120$

Rearrange: $50t - 70t = -120$ $-20t = -120$ $t = \frac{-120}{-20} = 6$

Step 5: Verify the solution

At $t = 6$ s:

• First ball: $s_1(6) = 50 \cdot 6 - 5 \cdot 6^2 = 300 - 180 = 120$ m
• Second ball: $s_2(6) = 50(6 - 2) - 5(6 - 2)^2 = 50 \cdot 4 - 5 \cdot 16 = 200 - 80 = 120$ m

Both displacements match, confirming the solution.

Common Traps & Exam Tip:

Trap 1: Students often confuse the time variable for the second ball. They mistakenly use $t$ instead of $t - 2$, leading to incorrect displacement equations.

Trap 2: Forgetting to expand $(t - 2)^2$ properly can result in sign errors. Always expand carefully: $(t - 2)^2 = t^2 - 4t + 4$

Exam Tip: When two objects are launched at different times, always define a common time variable (e.g., $t$ from the start) and express the motion of the second object relative to this common time. This avoids confusion and simplifies the algebra.