Equations Of Motion

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Stopping distance for particle in KN-010 is:

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25 m

20 m

15 m

10 m

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Subject: physics
Chapter: kinematics
Topic: equations of motion
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

25 m

From KN-010, $u = 10 m/s$ and $a = - 2 m/s^{2}$ . Using the third equation of motion: $v^{2} = u^{2} + 2 a s$ Since the particle stops, $v = 0$ : $0 = 1 0^{2} + 2 (- 2) s$ $4 s = 100 ⟹ s = 25 m$ . Alternatively, using $s = u t + \frac{1}{2} a t^{2}$ with $t = 5 s$ found in KN-010: $s = (10 \times 5) - \frac{1}{2} \times 2 \times 25 = 50 - 25 = 25 m$ .

This medium difficulty physics question is from the chapter kinematics, covering the topic of equations of motion. It appeared in the 2025 exam.

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Equations Of Motion

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