JEE PYQ: Motion in a Straight Line - Question ID de3b30f45622 (JEE Main 2018)

In problems involving braking and stopping distances, the key physics concept is kinematic motion under constant deceleration. When brakes are applied, the automobile experiences a constant negative acceleration (deceleration), bringing it to rest.

The fundamental kinematic equation used here is: $v^2 = u^2 + 2 a s$ where:

• $v$ = final velocity (0 m/s, since the vehicle stops),
• $u$ = initial velocity (in m/s),
• $a$ = acceleration (negative, since it's deceleration),
• $s$ = stopping distance.

Since the braking force (and hence deceleration) is assumed constant and the same in both cases, the stopping distance $s$ is proportional to the square of the initial speed. This is a crucial insight: doubling the speed quadruples the stopping distance.

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

Initial speed at 40 km/h: $40 \text{ km/h} = 40 \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{40000}{3600} = \frac{100}{9} \approx 11.11 \text{ m/s}$

Initial speed at 80 km/h: $80 \text{ km/h} = 80 \times \frac{1000}{3600} = \frac{80000}{3600} = \frac{200}{9} \approx 22.22 \text{ m/s}$

Step 2: Use kinematic equation to find deceleration

For the first case ($u_1 = \frac{100}{9}$ m/s, $s_1 = 40$ m, $v = 0$): $0 = \left(\frac{100}{9}\right)^2 + 2 a \cdot 40$ $\Rightarrow 2 a \cdot 40 = -\left(\frac{100}{9}\right)^2$ $\Rightarrow a = -\frac{1}{80} \left(\frac{100}{9}\right)^2$ We don’t need the exact value of $a$, only its constancy.

Step 3: Apply same deceleration to second case

For the second case ($u_2 = \frac{200}{9}$ m/s, same $a$, $v = 0$): $0 = \left(\frac{200}{9}\right)^2 + 2 a s_2$ $\Rightarrow 2 a s_2 = -\left(\frac{200}{9}\right)^2$

Step 4: Relate stopping distances using speed ratio

From Step 2 and Step 3: $\frac{s_2}{s_1} = \frac{\left(\frac{200}{9}\right)^2}{\left(\frac{100}{9}\right)^2} = \left(\frac{200}{100}\right)^2 = 4$ Thus: $s_2 = 4 \times s_1 = 4 \times 40 = 160 \text{ m}$

Students often mistakenly assume that stopping distance is directly proportional to speed (i.e., doubling speed doubles distance). This leads them to choose 80 m (Option B) instead of 160 m. The correct relationship is quadratic: stopping distance scales with the square of the speed.

Exam Tip: Always convert speeds to consistent units (preferably m/s) and use the kinematic equation $v^2 = u^2 + 2as$ for braking problems. Remember: no skidding implies constant deceleration, so the ratio of stopping distances equals the square of the speed ratio.