A boy moving with a velocity of 3 $ms^{-1}$ is brought to rest by a constant force after travelling 15m. Calculate the retardation.

Analyze the question: The boy starts at u = 3 m/s and ends at rest (v = 0) after covering s = 15 m. The force is constant, so the retardation is constant. We need its magnitude.

Concept: For constant acceleration, use the equation of motion that links velocity and distance without time:

v² = u² + 2as

Solution:

• u = 3 m/s
• v = 0
• s = 15 m

0 = (3)² + 2a(15)

0 = 9 + 30a

a = −9/30 = −0.30 m/s²

The negative sign shows the motion is slowing down, so the retardation is 0.30 m/s².

Cross-check: Average speed = (3 + 0)/2 = 1.5 m/s, so time = 15/1.5 = 10 s. Retardation = 3/10 = 0.30 m/s². This matches.

Tip: When a question gives velocities and distance but no time, use v² = u² + 2as straight away. "Retardation" just means negative acceleration, so give the answer as a positive value.