IBDP Maths SL 5.9 Kinematic problems involving displacement, velocity and acceleration AA HL Paper 2- Exam Style Questions- New Syllabus

(a) Maximum Speed
Speed is the magnitude of velocity, \( |v(t)| \). Using a graphing calculator to find the maximum of \( |e^{-\sin t} \cos(2t)| \) on the interval \( [0, 5] \):
The maximum value occurs at \( t \approx 3.32 \) seconds.
Max Speed \( \approx 2.296 \dots \)
Answer: \( \boxed{2.30 \, \text{m s}^{-1}} \) (to 3 s.f.)

(b) Total Distance Travelled
Total distance is given by the integral of speed:
Distance \( = \int_{0}^{5} |e^{-\sin t} \cos(2t)| \, dt \).
Using numerical integration on a calculator:
Distance \( \approx 3.9555 \dots \)
Answer: \( \boxed{3.96 \, \text{metres}} \) (to 3 s.f.)

(c) Acceleration at Second Change of Direction
A change in direction occurs when velocity \( v(t) = 0 \).
\( e^{-\sin t} \cos(2t) = 0 \implies \cos(2t) = 0 \).
1. First change: \( 2t = \frac{\pi}{2} \implies t_1 = \frac{\pi}{4} \approx 0.785 \).
2. Second change: \( 2t = \frac{3\pi}{2} \implies t_2 = \frac{3\pi}{4} \approx 2.35619 \dots \).
Acceleration is the derivative of velocity, \( a(t) = v'(t) \).
Using the product and chain rules:
\( v'(t) = (e^{-\sin t} \cdot -\cos t) \cos(2t) + e^{-\sin t} (-2 \sin(2t)) \).
At \( t = \frac{3\pi}{4} \), \( \cos(2t) = 0 \) and \( \sin(2t) = -1 \):
\( a(\frac{3\pi}{4}) = 0 + e^{-\sin(3\pi/4)} (-2(-1)) = 2e^{-1/\sqrt{2}} \approx 0.98613 \dots \)
Answer: \( \boxed{0.986 \, \text{m s}^{-2}} \) (to 3 s.f.)