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JEE PYQ: Motion in a Straight Line - Question ID 396c3480c03b (JEE Main 2022)

ID: 396c3480c03bJEE Main 2022Numerical Value

A car is moving with speed of $150 \mathrm{~km} / \mathrm{h}$ and after applying the break it will move $27 \mathrm{~m}$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ___________ m distance.

Your Answer

Step-by-step Explanation

Core Formula & Concept:

In problems involving braking and stopping distance, we rely on the kinematic equations of uniformly accelerated (or decelerated) motion. The key formula here is:

v^2 = u^2 + 2 a s

where

$v$ = final velocity (0 m/s when the car stops),
$u$ = initial velocity (converted to m/s),
$a$ = acceleration (negative, since it is deceleration),
$s$ = stopping distance.

The crucial insight is that the deceleration $a$ remains constant regardless of the initial speed. This allows us to relate stopping distances at different speeds.

Step-by-Step Derivation:

Step 1: Convert speeds to SI units

Initial speed

u_1 = 150 \text{ km/h} = 150 \times \frac{1000}{3600} = \frac{150 \times 5}{18} = \frac{125}{3} \text{ m/s} \approx 41.67 \text{ m/s}

Step 2: Apply the kinematic equation to find deceleration $a$

Given stopping distance

s_1 = 27 \text{ m}

and final velocity

v = 0

0 = u_1^2 + 2 a s_1 \implies a = -\frac{u_1^2}{2 s_1}

Substitute

u_1

and

s_1

a = -\frac{\left(\frac{125}{3}\right)^2}{2 \times 27} = -\frac{\frac{15625}{9}}{54} = -\frac{15625}{486} \text{ m/s}^2

Step 3: Determine the new initial speed

New speed

u_2 = \frac{1}{3} u_1 = \frac{1}{3} \times \frac{125}{3} = \frac{125}{9} \text{ m/s}

Step 4: Apply the same kinematic equation to find new stopping distance $s_2$

Using

v = 0

and the same

a

0 = u_2^2 + 2 a s_2 \implies s_2 = -\frac{u_2^2}{2 a}

Substitute

u_2

and

a

s_2 = -\frac{\left(\frac{125}{9}\right)^2}{2 \times \left(-\frac{15625}{486}\right)} = \frac{\frac{15625}{81}}{\frac{31250}{486}} = \frac{15625}{81} \times \frac{486}{31250}

Simplify:

s_2 = \frac{15625 \times 486}{81 \times 31250} = \frac{15625 \times 6}{31250} = \frac{93750}{31250} = 3 \text{ m}

Step 5: Conclusion

The car stops after travelling 3 m when moving at one-third the original speed. Common Traps & Exam Tip:

Students often forget to convert speeds from km/h to m/s, leading to incorrect deceleration values. Another common mistake is assuming stopping distance scales linearly with speed—it actually scales with the square of the speed ( $s \propto u^2$ ). This quadratic relationship is the key to solving such problems efficiently.

Exam Tip: Always check units and remember that stopping distance depends on the square of the initial velocity when deceleration is constant.

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