JEE PYQ: Motion in a Straight Line - Question ID 46b1612c76a7 (JEE Main 2005)

Core Formula & Concept:

In problems involving motion in a straight line with varying acceleration, we break the journey into distinct phases and apply the kinematic equations for each phase. The key formulas used are:

• For uniformly accelerated motion (starting from rest): $v = u + at$ $s = ut + \frac{1}{2} a t^2$ $v^2 = u^2 + 2 a s$ Here, $u = 0$, so these simplify to: $v = a t$ $s = \frac{1}{2} a t^2$ $v^2 = 2 a s$
• For motion at constant speed: distance = speed × time, i.e., $s = v t$.
• For uniformly decelerated motion (coming to rest): We use the same kinematic equations but with negative acceleration. The final velocity is zero.

The total distance traversed is the sum of distances covered in all phases. We are given that this total distance is $15 S$, where $S$ is the distance covered during the initial acceleration phase.

Step-by-Step Derivation:

Phase 1: Acceleration from rest at rate $f$ for distance $S$

Let $t_1$ be the time taken to cover distance $S$ under acceleration $f$. Since the car starts from rest:

$S = \frac{1}{2} f t_1^2 \quad \text{(1)}$

The velocity at the end of this phase (which is also the constant speed in the next phase) is:

$v = f t_1 \quad \text{(2)}$

Phase 2: Motion at constant speed $v$ for time $t$

The distance covered in this phase is:

$S_2 = v t = f t_1 t \quad \text{(3)}$

Phase 3: Deceleration at rate $\frac{f}{2}$ to come to rest

Let $t_3$ be the time taken to decelerate to rest. Using $v = u + a t$ with $v = 0$, $u = v$, and $a = -\frac{f}{2}$:

$0 = v - \frac{f}{2} t_3$ $t_3 = \frac{2 v}{f} = \frac{2 f t_1}{f} = 2 t_1 \quad \text{(4)}$

The distance covered during deceleration is:

$S_3 = v t_3 - \frac{1}{2} \frac{f}{2} t_3^2$ Substituting $v = f t_1$ and $t_3 = 2 t_1$: $S_3 = f t_1 (2 t_1) - \frac{1}{4} f (2 t_1)^2$ $S_3 = 2 f t_1^2 - \frac{1}{4} f (4 t_1^2)$ $S_3 = 2 f t_1^2 - f t_1^2 = f t_1^2 \quad \text{(5)}$

Total distance traversed:

The total distance is the sum of distances in all three phases:

$S_{\text{total}} = S + S_2 + S_3 = S + f t_1 t + f t_1^2$ We are given $S_{\text{total}} = 15 S$. Substituting: $S + f t_1 t + f t_1^2 = 15 S$ $f t_1 t + f t_1^2 = 14 S$ $f t_1 (t + t_1) = 14 S \quad \text{(6)}$

Expressing $t_1$ in terms of $S$:

From equation (1): $S = \frac{1}{2} f t_1^2$ $t_1^2 = \frac{2 S}{f}$ $t_1 = \sqrt{\frac{2 S}{f}} \quad \text{(7)}$

Substituting $t_1$ into equation (6):

$f \sqrt{\frac{2 S}{f}} \left( t + \sqrt{\frac{2 S}{f}} \right) = 14 S$ Simplify $f \sqrt{\frac{2 S}{f}}$: $f \sqrt{\frac{2 S}{f}} = \sqrt{f^2 \cdot \frac{2 S}{f}} = \sqrt{2 f S}$ So the equation becomes: $\sqrt{2 f S} \left( t + \sqrt{\frac{2 S}{f}} \right) = 14 S$ Let $\sqrt{\frac{2 S}{f}} = x$. Then $\sqrt{2 f S} = f x$, and the equation becomes: $f x (t + x) = 14 S$ But $x^2 = \frac{2 S}{f}$, so $S = \frac{f x^2}{2}$. Substitute $S$: $f x (t + x) = 14 \cdot \frac{f x^2}{2}$ $f x (t + x) = 7 f x^2$ Divide both sides by $f x$ (assuming $f \neq 0$, $x \neq 0$): $t + x = 7 x$ $t = 6 x$ Recall $x = \sqrt{\frac{2 S}{f}}$, so: $t = 6 \sqrt{\frac{2 S}{f}}$ Square both sides: $t^2 = 36 \cdot \frac{2 S}{f}$ $t^2 = \frac{72 S}{f}$ $S = \frac{1}{72} f t^2$

This matches option D.

Common Traps & Exam Tip:

Students often make the following mistakes in this problem:

• Incorrectly calculating the deceleration distance: Many forget that the deceleration phase uses $\frac{f}{2}$ and incorrectly apply the kinematic equations, leading to wrong expressions for $S_3$. Always verify the sign of acceleration and the final velocity.
• Miscounting the total distance: Some students add the distances incorrectly or misinterpret the given total distance as $15 S$, forgetting that $S$ is only the distance covered during the first phase.
• Algebraic errors in substitution: The substitution step involving $\sqrt{\frac{2 S}{f}}$ is tricky. Students often make errors in simplifying $f \sqrt{\frac{2 S}{f}}$ or in handling the square roots. Always simplify expressions step-by-step.
• Assuming time symmetry: Some assume that the time taken to accelerate is equal to the time taken to decelerate, which is not true here because the deceleration is half the acceleration. Always derive the times using kinematic equations.

Exam Tip: Break the problem into clear phases, label all variables, and verify each step. For problems involving multiple phases, it is often helpful to express all distances and times in terms of a single variable (here, $t_1$) before substituting into the total distance equation.