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QPage 48-49: Q10.1: HC Verma: Concepts of Physics - Volume 1 Chapter-3 Rest and Motion Kinematics Page 48-49: Q10.1 Solution: Give example where the velocity of a particle is zero but its…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often equate a momentary zero speed with zero acceleration, overlooking that acceleration is the time‑derivative of velocity and can be non‑zero at turning points where forces act (e.g., gravity on a thrown ball). Always examine the net force or the derivative of the velocity function, not just the instantaneous speed.

Give example where the velocity of a particle is zero but its acceleration is not ...

Step-by-Step Verified Solution

Classic Example 1: A Ball Thrown Vertically Upward

  1. Throw a ball straight up with some initial speed.
  2. It rises, slows down, and at the highest point of its trajectory:
  3. Velocity (v) = 0 (it momentarily stops before falling back down).
  4. Acceleration (a) = –g = –9.8 m/s² (gravity is still pulling it downward, and this acceleration is constant throughout the motion).

At the top:

v = 0, but a = –9.8 m/s² ≠ 0

Classic Example 2: A Simple Pendulum

  1. A pendulum bob swings back and forth.
  2. At the extreme ends of its swing (the leftmost or rightmost point):
  3. Velocity (v) = 0 (it momentarily stops before reversing direction).
  4. Acceleration (a) ≠ 0 (directed towards the equilibrium position, due to the restoring force of gravity).

At the extreme position:

v = 0, but a = (g/L) × displacement ≠ 0

Classic Example 3: A Particle in Simple Harmonic Motion (SHM)

  1. Consider a mass attached to a spring oscillating horizontally.
  2. At the maximum displacement (amplitude, x = ±A):
  3. Velocity (v) = 0 (it stops before turning back).
  4. Acceleration (a) = –ω²x = –ω²A ≠ 0 (maximum magnitude, directed towards the mean position).

At the extreme point:

v = 0, but a = –ω²A ≠ 0

Classic Example 4: A Car at the Instant of Braking to a Stop

  1. A car is moving forward and applies brakes.
  2. Just at the exact instant when the car comes to a complete stop (v = 0):
  3. If the brakes are still applied, the car still has deceleration (negative acceleration) because the braking force is still acting.
  4. So v = 0, but a ≠ 0 (until the brakes are released).


All Questions in this Chapter

QP 49-50: Q1 QP 49-50: Q2 QP 49-50: Q3 QP 49-50: Q4 QP 49-50: Q5 QP 49-50: Q6 QP 49-50: Q7 QP 49-50: Q8 QP 49-50: Q9 QP 49-50: Q10 QP 49-50: Q11 QP 49-50: Q12 QP 49-50: Q13 QP 50-51: Q1 QP 50-51: Q2 QP 50-51: Q3 QP 50-51: Q4 QP 50-51: Q5 QP 50-51: Q6 QP 50-51: Q7 QP 50-51: Q8 QP 50-51: Q9 QP 50-51: Q10 QP 51-54: Q1 QP 51-54: Q2 QP 51-54: Q3.1 QP 51-54: Q3.2 QP 51-54: Q3.3 QP 51-54: Q3.4 QP 51-54: Q4.1 QP 51-54: Q4.2 QP 51-54: Q5 QP 51-54: Q6 QP 51-54: Q7 QP 51-54: Q8.1 QP 51-54: Q8.2 QP 51-54: Q8.3 QP 51-54: Q9.1 QP 51-54: Q9.2 QP 51-54: Q10 QP 51-54: Q11 QP 51-54: Q12 QP 51-54: Q13 QP 51-54: Q14 QP 51-54: Q15.1 QP 51-54: Q15.2 QP 51-54: Q15.3 QP 51-54: Q16 QP 51-54: Q17 QP 51-54: Q18.1 QP 51-54: Q18.2 QP 51-54: Q19 QP 51-54: Q20 QP 51-54: Q21 QP 51-54: Q22 QP 51-54: Q23.1 QP 51-54: Q23.2 QP 51-54: Q23.3 QP 51-54: Q24 QP 51-54: Q25.1 QP 51-54: Q25.2 QP 51-54: Q25.3 QP 51-54: Q26 QP 51-54: Q27 QP 51-54: Q28 QP 51-54: Q29 QP 51-54: Q30 QP 51-54: Q31 QP 51-54: Q32.1 QP 51-54: Q32.2 QP 51-54: Q32.3 QP 51-54: Q33.1 QP 51-54: Q33.2 QP 51-54: Q34 QP 51-54: Q35 QP 51-54: Q36 QP 51-54: Q37 QP 51-54: Q38 QP 51-54: Q39 QP 51-54: Q40 QP 51-54: Q41 QP 51-54: Q42 QP 51-54: Q43.1 QP 51-54: Q43.2 QP 51-54: Q44 QP 51-54: Q45 QP 51-54: Q46.1 QP 51-54: Q46.2 QP 51-54: Q47.1 QP 51-54: Q47.2 QP 51-54: Q48 QP 51-54: Q49.1 QP 51-54: Q49.2 QP 51-54: Q50 QP 51-54: Q51 QP 51-54: Q52 QPage 48-49: Q1 QPage 48-49: Q2 QPage 48-49: Q3 QPage 48-49: Q4 QPage 48-49: Q5 QPage 48-49: Q6 QPage 48-49: Q7 QPage 48-49: Q8 QPage 48-49: Q9 QPage 48-49: Q10.1 QPage 48-49: Q10.2 QPage 48-49: Q10.3 QPage 48-49: Q11 QPage 48-49: Q12 QPage 48-49: Q13 QPage 48-49: Q14

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