Solutions — Displacement, Velocity and Motion Graphs

Q1 — x = t² − 4t + 3

(a) v(t) = x′(t) = 2t − 4

(b) a(t) = v′(t) = 2 m/s² (constant)

(c) At t = 3: v(3) = 2(3) − 4 = 2 m/s (moving in positive direction). a(3) = 2 m/s².

Q2 — x = t³ − 6t² + 9t

(a) x(0) = 0 cm (starts at origin). v(t) = 3t² − 12t + 9; v(0) = 9 cm/s.

(b) v(t) = 3(t² − 4t + 3) = 3(t − 1)(t − 3) = 0 → t = 1 s and t = 3 s.

(c) x(1) = 1 − 6 + 9 = 4 cm. x(3) = 27 − 54 + 27 = 0 cm.

Q3 — x = t² − 2t − 3

(a) v(t) = 2t − 2

(b) v < 0 when 2t − 2 < 0 → t < 1. Particle moves in negative direction for 0 ≤ t < 1.

(c) v > 0 when t > 1. Particle moves in positive direction for t > 1.

Q4 — x = 2t³ − 9t²

v(t) = 6t² − 18t = 6t(t − 3)

(a) v = 0 at t = 0 and t = 3.

(b) Sign of v: t < 0: v > 0 (not relevant, t ≥ 0); 0 < t < 3: v = 6t(t−3) < 0 (moving left); t > 3: v > 0 (moving right).
At t = 0: v changes from 0 to negative → particle starts moving left. Direction changes at t = 3 s.

(c) x(0) = 0 m. x(3) = 2(27) − 9(9) = 54 − 81 = −27 m.

Q5 — Ball: h = 20t − 5t²

v(t) = h′(t) = 20 − 10t.

(a) Maximum height when v = 0: 20 − 10t = 0 → t = 2. h(2) = 40 − 20 = 20 metres.

(b) Ball hits ground when h = 0: 20t − 5t² = 0 → 5t(4 − t) = 0 → t = 0 or t = 4. Hits ground at t = 4 s.

(c) v(4) = 20 − 40 = −20 m/s. The negative sign indicates downward motion — the ball is falling when it hits the ground.

Q6 — v = 6t² − 4t

(a) a(t) = v′(t) = 12t − 4

(b) a(2) = 12(2) − 4 = 20 m/s²

(c) v = 0: 6t² − 4t = 2t(3t − 2) = 0 → t = 0 or t = 2⁄3 s.

Q7 — x = t² − 4t from t = 0 to t = 4

(a) v(t) = 2t − 4 = 0 → t = 2. Particle changes direction at t = 2 s.

(b) x(0) = 0; x(2) = 4 − 8 = −4; x(4) = 16 − 16 = 0.
Distance 0 → 2: |−4 − 0| = 4; Distance 2 → 4: |0 − (−4)| = 4.
Total distance = 8 metres.

(c) Displacement = x(4) − x(0) = 0 − 0 = 0 metres.

(d) The particle moved 4 m to the left then 4 m back, ending where it started. Displacement = 0, but it actually travelled 8 m in total.

Q8 — Two particles: x_A = t² + 2t, x_B = 3t

(a) x_A = x_B: t² + 2t = 3t → t² − t = 0 → t(t − 1) = 0 → t = 0 or t = 1.
They meet again at t = 1 s. Position: x_B(1) = 3 m.

(b) v_A(t) = 2t + 2 → v_A(1) = 4 m/s.
v_B(t) = 3 → v_B(1) = 3 m/s (constant).

(c) At t = 1 s, particle A (4 m/s) is moving faster than particle B (3 m/s).

Q9 — x = t³ − 3t² + 3t + 1

(a) v(t) = 3t² − 6t + 3 = 3(t² − 2t + 1) = 3(t − 1)²

(b) (t − 1)² ≥ 0 for all real t, so v(t) = 3(t − 1)² ≥ 0 always. The velocity is never negative, so the particle never moves in the negative direction and therefore never reverses. (At t = 1 it momentarily stops but then continues in the same direction.)

(c) a(t) = v′(t) = 6t − 6. a = 0 when 6t − 6 = 0 → t = 1 s. This is when the particle is momentarily at rest and acceleration is zero.

Q10 — Reading a displacement-time graph

(a) Velocity is zero where the displacement-time graph has a horizontal tangent (turning point). The graph has a minimum at t = 3 (where x = −1). So v = 0 at approximately t = 3.

(b) Velocity is negative where the graph is decreasing. The graph decreases from t = 0 to t = 3 (x goes from 3 to −1). So velocity is negative on 0 < t < 3.

(c) Speed |v| is greatest where the graph is steepest. This appears to be near t = 0 (steep initial drop) or near t = 5–6 (steep rise back to 3). The greatest steepness determines greatest speed.

(d) The particle moves from x = 3 to x = −1 (a drop of 4 units), then returns from x = −1 to x = 3 (a rise of 4 units). Total distance = 4 + 4 = 8 units.