0606/22 May/Jun 2025, Question 2: Drawing Straight-Line Graphs

Part (a) · 2 marks

Sketch the displacement–time graph and state where it crosses the t- and s-axes.

1. 1s = 0 when t = 1 and t = 4 (repeated root, so the curve touches the t-axis at t = 4)
2. 2When t = 0, s = (−4)²(−1) = −16
3. 3Positive cubic: rises through t = 1 to a maximum, falls to touch the axis at t = 4, then rises

AnswerCubic for t ≥ 0 starting at (0, −16), crossing the t-axis at t = 1, maximum in the first quadrant, touching the t-axis at the minimum t = 4, then increasing

Part (b) · 2 marks

Differentiate s to obtain v, giving the result factorised.

1. 1s = t³ − 9t² + 24t − 16
2. 2v = ds/dt = 3t² − 18t + 24
3. 3v = 3(t² − 6t + 8) = 3(t − 4)(t − 2)

Answerv = 3(t − 4)(t − 2)

Part (c) · 2 marks

Sketch the velocity–time graph and state its axis intercepts.

1. 1v = 3(t − 4)(t − 2): t-intercepts t = 2 and t = 4
2. 2When t = 0, v = 24
3. 3Minimum at t = 3, v = −3 (below the t-axis)

AnswerU-shaped parabola for t ≥ 0 with v-intercept 24, crossing the t-axis at t = 2 and t = 4, minimum in the fourth quadrant

Part (d) · 1 mark

Differentiate v to obtain an expression for the acceleration a.

1. 1a = dv/dt = d/dt(3t² − 18t + 24)
2. 2a = 6t − 18

Answera = 6t − 18

Part (e) · 3 marks

Sketch the acceleration–time graph and state its axis intercepts.

1. 1a = 6t − 18 is a straight line with gradient 6
2. 2When t = 0, a = −18
3. 3a = 0 when t = 3

AnswerStraight line with positive gradient, a-intercept −18, crossing the t-axis at t = 3

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