IB Mathematics AHL 5.13 Kinematic problems involving displacement-AI HL Paper 1- Exam Style Questions- New Syllabus

(a)

(i) The graph of \( v(t) = 2\cos t + \sin 2t – 0.2 \) over \( 0 \leq t \leq 6 \) is a periodic‑type curve starting at \( v(0) = 2 + 0 – 0.2 = 1.8 \). It oscillates with two local maxima and one local minimum in the interval, crossing the \( t\)-axis where the velocity changes sign. A correct sketch should show the general shape with the correct starting point and behaviour within the given domain.

(ii) Acceleration is zero when \( a(t) = v'(t) = 0 \). \( v'(t) = -2\sin t + 2\cos 2t \). Solving \( -2\sin t + 2\cos 2t = 0 \) gives critical points at approximately:
\( t \approx 0.524 \) s, \( v \approx 2.40 \) m s\(^{-1}\) – local maximum.
\( t \approx 2.62 \) s, \( v \approx -2.80 \) m s\(^{-1}\) – local minimum.
\( t \approx 4.71 \) s, \( v \approx -0.20 \) m s\(^{-1}\) – stationary point (inflection type). These points should be labelled on the sketch with coordinates (0.524, 2.40), (2.62, –2.80), (4.71, –0.20).

(b)

(i) Distance travelled = total area between the velocity curve and the \( t\)-axis, taking absolute values of \( v(t) \) when it is negative.

Integral expression:
\( \int_{0}^{6} |v(t)| \, dt \).

Alternatively, because \( v(t) \) changes sign, it can be written as the sum of separate integrals over intervals where \( v(t) \) does not change sign: \( \int_{0}^{1.52074} v(t) \, dt – \int_{1.52074}^{5.31535} v(t) \, dt + \int_{5.31535}^{6} v(t) \, dt \).

(ii) Using a GDC to evaluate the integral of \( |v(t)| \) over \( 0 \le t \le 6 \):
Distance \( \approx 7.76 \) m (7.76487… m).

✅ Answer: \( \boxed{7.76 \text{ m}} \) (to 3 s.f.)