JEE PYQ: Motion in a Straight Line - Question ID 5b00b7944e92 (JEE Main 2004)

When a vehicle brakes, it decelerates uniformly until it comes to rest. The key physics principle here is the kinematic equation for uniformly accelerated (or decelerated) motion. Specifically, we use the equation that relates initial velocity ($u$), final velocity ($v$), acceleration ($a$), and displacement ($s$):

In braking scenarios:

• The final velocity $v = 0$ (since the vehicle stops).
• The acceleration $a$ is negative (deceleration) and constant.
• The stopping distance $s$ is the displacement during braking.

This shows that the stopping distance $s$ is proportional to the square of the initial velocity ($s \propto u^2$), assuming the braking deceleration $a$ remains constant.

Step 1: Convert speeds to consistent units (m/s)

Given:

• Initial speed $u_1 = 60 \text{ km/h}$
• Stopping distance $s_1 = 20 \text{ m}$

Step 2: Calculate the deceleration ($a$)

Using the kinematic equation for the first scenario: $v^2 = u_1^2 + 2as_1$ Since $v = 0$: $0 = \left(\frac{50}{3}\right)^2 + 2a(20)$ $0 = \frac{2500}{9} + 40a$ $40a = -\frac{2500}{9}$ $a = -\frac{2500}{9 \times 40} = -\frac{2500}{360} = -\frac{250}{36} = -\frac{125}{18} \text{ m/s}^2$

Step 3: Apply the same deceleration to the new speed

New speed $u_2 = 120 \text{ km/h}$ (twice the original speed). Convert to m/s: $u_2 = 120 \times \frac{1000}{3600} = \frac{120000}{3600} = \frac{100}{3} \text{ m/s} \approx 33.33 \text{ m/s}$ Using the same kinematic equation for the new scenario: $0 = u_2^2 + 2as_2$ $0 = \left(\frac{100}{3}\right)^2 + 2 \left(-\frac{125}{18}\right) s_2$ $0 = \frac{10000}{9} - \frac{250}{18} s_2$ Multiply through by 18 to eliminate denominators: $0 = 20000 - 250 s_2$ $250 s_2 = 20000$ $s_2 = \frac{20000}{250} = 80 \text{ m}$

Alternative (Simpler) Approach: Using Proportionality

Since $s \propto u^2$ (from $s = -\frac{u^2}{2a}$), if the speed doubles ($u_2 = 2u_1$), the stopping distance becomes: $s_2 = (2)^2 s_1 = 4 \times 20 \text{ m} = 80 \text{ m}$ This confirms our earlier result. Common Traps & Exam Tip:

Trap 1: Linear Thinking
Many students mistakenly assume that if speed doubles, stopping distance also doubles (linear relationship). This is incorrect because stopping distance depends on the square of the speed. Always remember: $s \propto u^2$.

Trap 2: Unit Inconsistency
Failing to convert km/h to m/s before calculations leads to incorrect results. Always ensure consistent units (preferably SI units like m/s for speed and m for distance).

Trap 3: Ignoring Negative Sign for Deceleration
The acceleration $a$ is negative during braking, but students sometimes drop the negative sign, leading to incorrect calculations. Always verify the sign of $a$ based on the context (deceleration = negative acceleration).

Exam Tip:
For such problems, the proportionality method ($s \propto u^2$) is faster and reduces calculation errors. Use it to cross-verify your answer. If the speed becomes $n$ times the original, the stopping distance becomes $n^2$ times the original. Here, $n = 2 \implies s_2 = 4 \times 20 = 80 \text{ m}$.