The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is?

Analyze the question: A particle starts from rest (u = 0) and moves in a straight line. We need the relationship between displacement s and time t. The question doesn't say the acceleration is constant, but the standard assumption for this type of question is uniform acceleration.

Concept: For uniform acceleration, the equation of motion is

s = ut + ½at²

Solution:

1. The particle starts from rest, so u = 0.
2. Substitute into the equation:

s = (0)t + ½at² = ½at²

3. Since ½ and a are constants (uniform acceleration), we can write:

s ∝ t²

So displacement is directly proportional to the square of time. This means if time doubles, displacement becomes four times as large.

Checking the other options:

• A (t⁻²) and B (t⁻¹): these would mean displacement decreases as time increases, which is wrong for a particle moving away from rest.
• C (t): this applies to motion at constant velocity (s = vt), but our particle starts from rest and so must accelerate.
• D (t²): matches s = ½at².

Answer: D. s ∝ t²

Tip: When you see "initially at rest," set u = 0 immediately. Then s = ½at² (s ∝ t²), v = at (v ∝ t), and v² = 2as (v² ∝ s). A graph of s against t² will be a straight line through the origin, and this is a common WASSCE graph question.