JEE PYQ: Motion in a Straight Line - Question ID e1d5bb605b90 (JEE Main 2019)

Core Formula & Concept:

In problems involving motion with constant acceleration, the following kinematic relations are fundamental:

• Displacement ($s$) as a function of time ($t$) and acceleration ($a$): $s = ut + \tfrac{1}{2} a t^2$ where $u$ is the initial velocity.
• Final velocity ($v$) as a function of initial velocity, acceleration, and time: $v = u + a t$
• Relation between displacement, initial velocity, final velocity, and acceleration (without explicit time): $v^2 = u^2 + 2 a s$

Since both cars start from rest, their initial velocities are zero ($u = 0$). The key idea is to express the race distance in terms of each car’s acceleration and travel time, then relate the time difference and speed difference at the finish.

Step-by-Step Derivation:

Let:

• $a_1$ = acceleration of car A
• $a_2$ = acceleration of car B
• $t_A$ = time taken by car A to finish
• $t_B$ = time taken by car B to finish
• $v_A$ = speed of car A at finish
• $v_B$ = speed of car B at finish

1. Express the race distance $s$ in two ways:

Since both cars cover the same distance $s$ from rest:

$s = \tfrac{1}{2} a_1 t_A^2 \quad \text{(for car A)}$ $s = \tfrac{1}{2} a_2 t_B^2 \quad \text{(for car B)}$

Equate the two expressions:

$\tfrac{1}{2} a_1 t_A^2 = \tfrac{1}{2} a_2 t_B^2$ $\Rightarrow a_1 t_A^2 = a_2 t_B^2 \quad \text{(1)}$

2. Relate the time difference:

Car A finishes $t$ seconds earlier than car B:

$t_B = t_A + t \quad \text{(2)}$

3. Express the speed difference at finish:

Since both cars start from rest, their final speeds are:

$v_A = a_1 t_A$ $v_B = a_2 t_B$

The problem states that car A passes the finish with speed $v$ more than car B:

$v_A = v_B + v$ $\Rightarrow a_1 t_A = a_2 t_B + v \quad \text{(3)}$

4. Substitute $t_B$ from (2) into (1):

$a_1 t_A^2 = a_2 (t_A + t)^2$ Expand the right side: $a_1 t_A^2 = a_2 (t_A^2 + 2 t_A t + t^2)$ $a_1 t_A^2 = a_2 t_A^2 + 2 a_2 t_A t + a_2 t^2$ Bring all terms to one side: $(a_1 - a_2) t_A^2 - 2 a_2 t t_A - a_2 t^2 = 0$

This is a quadratic in $t_A$. However, instead of solving for $t_A$, we use equation (3) to eliminate $t_A$.

5. Express $v$ in terms of $t_A$ and $t$:

From (3): $v = a_1 t_A - a_2 t_B$ Substitute $t_B = t_A + t$: $v = a_1 t_A - a_2 (t_A + t)$ $v = (a_1 - a_2) t_A - a_2 t \quad \text{(4)}$

6. Solve for $t_A$ from equation (1):

From (1): $a_1 t_A^2 = a_2 t_B^2$ $\Rightarrow t_A^2 = \frac{a_2}{a_1} t_B^2$ $\Rightarrow t_A = \sqrt{\frac{a_2}{a_1}} t_B$ Substitute $t_B = t_A + t$: $t_A = \sqrt{\frac{a_2}{a_1}} (t_A + t)$ Square both sides: $t_A^2 = \frac{a_2}{a_1} (t_A + t)^2$ $a_1 t_A^2 = a_2 (t_A + t)^2$ This brings us back to the earlier quadratic. Instead, let’s use a substitution to simplify.

7. Introduce a ratio variable:

Let $k = \frac{t_A}{t}$. Then $t_A = k t$, and from (2): $t_B = t_A + t = k t + t = (k + 1) t$ Substitute into (1): $a_1 (k t)^2 = a_2 ((k + 1) t)^2$ $a_1 k^2 t^2 = a_2 (k + 1)^2 t^2$ Cancel $t^2$: $a_1 k^2 = a_2 (k + 1)^2$ $\Rightarrow \frac{a_1}{a_2} = \frac{(k + 1)^2}{k^2}$ Take square roots (assuming positive accelerations and times): $\sqrt{\frac{a_1}{a_2}} = \frac{k + 1}{k}$ $\Rightarrow k \sqrt{\frac{a_1}{a_2}} = k + 1$ $\Rightarrow k \left( \sqrt{\frac{a_1}{a_2}} - 1 \right) = 1$ $\Rightarrow k = \frac{1}{\sqrt{\frac{a_1}{a_2}} - 1} = \frac{\sqrt{a_2}}{\sqrt{a_1} - \sqrt{a_2}}$

8. Express $v$ in terms of $k$ and $t$:

From (4): $v = (a_1 - a_2) t_A - a_2 t$ Substitute $t_A = k t$: $v = (a_1 - a_2) k t - a_2 t$ $v = t \left[ (a_1 - a_2) k - a_2 \right]$ Substitute $k = \frac{\sqrt{a_2}}{\sqrt{a_1} - \sqrt{a_2}}$: $v = t \left[ (a_1 - a_2) \frac{\sqrt{a_2}}{\sqrt{a_1} - \sqrt{a_2}} - a_2 \right]$ Simplify the numerator: $(a_1 - a_2) = (\sqrt{a_1} - \sqrt{a_2})(\sqrt{a_1} + \sqrt{a_2})$ Thus: $(a_1 - a_2) \frac{\sqrt{a_2}}{\sqrt{a_1} - \sqrt{a_2}} = (\sqrt{a_1} + \sqrt{a_2}) \sqrt{a_2}$ So: $v = t \left[ (\sqrt{a_1} + \sqrt{a_2}) \sqrt{a_2} - a_2 \right]$ $v = t \left[ \sqrt{a_1 a_2} + a_2 - a_2 \right]$ $v = t \sqrt{a_1 a_2}$

Thus, the correct expression for $v$ is:

$v = \sqrt{a_1 a_2} t$ Common Traps & Exam Tip:

Students often make these mistakes:

• Incorrectly relating time and speed: Many assume that the speed difference is directly proportional to the acceleration difference, ignoring the quadratic relation between time and distance.
• Algebraic errors in substitution: The quadratic in $t_A$ can be messy; substituting a ratio variable (like $k = t_A / t$) simplifies the algebra.
• Misapplying kinematic equations: Forgetting that both cars start from rest and using incorrect initial conditions leads to wrong expressions.
• Overcomplicating the problem: Some students try to solve for $t_A$ explicitly, which is unnecessary. The key is to express everything in terms of the given time difference $t$.

Exam Tip: Always start by writing down known quantities and relations. Use symmetry and substitution to reduce variables. In this problem, introducing $k = t_A / t$ simplifies the algebra significantly.