JEE PYQ: Motion in a Straight Line - Question ID 6f21fa5c9d50 (JEE Main 2003)

ID: 6f21fa5c9d50JEE Main 2003Single Correct MCQ
A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

Select Option

Step-by-step Explanation

Core Formula & Concept:

In problems involving braking distances, the key physics concept is the work-energy theorem. When brakes are applied, the kinetic energy of the car is dissipated by the work done by the braking force. Assuming the braking force is constant, the work done by the brakes equals the change in kinetic energy of the car.

The relevant formulas are:

  • Kinetic energy: K=12mv2K = \frac{1}{2} m v^2, where mm is the mass of the car and vv is its speed.
  • Work done by braking force: W=FdW = F \cdot d, where FF is the magnitude of the braking force and dd is the stopping distance.

Since the car comes to rest, the work done by the brakes equals the initial kinetic energy: Fd=12mv2F \cdot d = \frac{1}{2} m v^2

From this, we can express the stopping distance dd as: d=mv22Fd = \frac{m v^2}{2 F}

Notice that the stopping distance depends on the square of the speed. This is the crucial insight for solving the problem.

Step-by-Step Derivation:

Step 1: Convert speeds to consistent units
The given speeds are in km/hr. To make calculations easier, convert them to m/s:

50 km/hr=50×1000 m3600 s=500003600=125913.89 m/s50 \text{ km/hr} = 50 \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{50000}{3600} = \frac{125}{9} \approx 13.89 \text{ m/s} 100 km/hr=100×1000 m3600 s=1000003600=250927.78 m/s100 \text{ km/hr} = 100 \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{100000}{3600} = \frac{250}{9} \approx 27.78 \text{ m/s}

However, since we are dealing with ratios, we can avoid explicit conversion by working directly with the ratio of speeds.

Step 2: Express stopping distance in terms of speed
From the work-energy relation: d=mv22Fd = \frac{m v^2}{2 F}

Let d1d_1 be the stopping distance at speed v1=50 km/hrv_1 = 50 \text{ km/hr}, and d2d_2 be the stopping distance at speed v2=100 km/hrv_2 = 100 \text{ km/hr}. Then:

d1=mv122F,d2=mv222Fd_1 = \frac{m v_1^2}{2 F}, \quad d_2 = \frac{m v_2^2}{2 F}

Taking the ratio of d2d_2 to d1d_1:

d2d1=mv222Fmv122F=v22v12=(v2v1)2\frac{d_2}{d_1} = \frac{\frac{m v_2^2}{2 F}}{\frac{m v_1^2}{2 F}} = \frac{v_2^2}{v_1^2} = \left( \frac{v_2}{v_1} \right)^2

Step 3: Substitute the given values
Given v1=50 km/hrv_1 = 50 \text{ km/hr}, v2=100 km/hrv_2 = 100 \text{ km/hr}, and d1=6 md_1 = 6 \text{ m}:

d26=(10050)2=22=4\frac{d_2}{6} = \left( \frac{100}{50} \right)^2 = 2^2 = 4

Thus:

d2=4×6=24 md_2 = 4 \times 6 = 24 \text{ m}

Conclusion: The minimum stopping distance at 100 km/hr is 24 m, which corresponds to option C.

Common Traps & Exam Tip:

Trap 1: Linear thinking
Many students mistakenly assume that doubling the speed doubles the stopping distance. This leads them to choose 12 m (option A). However, stopping distance depends on the square of the speed, not the speed itself.

Trap 2: Ignoring unit consistency
While the ratio method avoids explicit unit conversion, some students convert speeds incorrectly, leading to calculation errors. Always ensure units are consistent when performing numerical calculations.

Exam Tip:
When dealing with braking or stopping distances, always check if the relationship is linear or quadratic with respect to speed. The work-energy theorem is a powerful tool for such problems, and recognizing the v2v^2 dependence is key to avoiding mistakes.

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