IB Mathematics SL 5.9 Kinematic problems AA SL Paper 2- Exam Style Questions- New Syllabus

(a)
Maximum speed is the maximum value of \( |v(t)| \) on \( 0 \leq t \leq 5 \).
Using a GDC to graph \( y = |e^{-\sin x} \cos(2x)| \), the maximum occurs at approximately \( t \approx 4.71238 \) ( \( = \frac{3\pi}{2} \) ).
At this point, \( \sin t = -1 \), so \( e^{-\sin t} = e^{1} = e \), and \( \cos(2t) = \cos(3\pi) = -1 \).
Thus \( v = e \times (-1) = -e \), so speed \( = |v| = e \).
\( \boxed{2.72} \ \text{ms}^{-1} \) (or \( e \ \text{ms}^{-1} \)).

(b)
Total distance travelled is \( \int_{0}^{5} |v(t)| \, dt \).
Using a GDC: \( \int_{0}^{5} |e^{-\sin t} \cos(2t)| \, dt \approx 3.84591 \).
\( \boxed{3.85} \ \text{m} \).

(c)
The particle reverses direction when \( v(t) = 0 \).
Solve \( e^{-\sin t} \cos(2t) = 0 \implies \cos(2t) = 0 \).
Within \( 0 \leq t \leq 5 \), the zeros occur at \( 2t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \dots \) so \( t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4} \dots \).
The second reversal occurs at \( t = \frac{3\pi}{4} \) (\( \approx 2.35619 \)).
Acceleration \( a(t) = v'(t) \). Using the product and chain rules:
\( a(t) = -e^{-\sin t} \cos t \cos(2t) – 2e^{-\sin t} \sin(2t) \).
At \( t = \frac{3\pi}{4} \): \( \sin t = \frac{\sqrt{2}}{2}, \cos t = -\frac{\sqrt{2}}{2}, \sin(2t) = -1, \cos(2t) = 0 \).
Since \( \cos(2t) = 0 \), the first term is zero:
\( a = -2e^{-\sin(\frac{3\pi}{4})} \sin(\frac{3\pi}{2}) = -2e^{-\frac{\sqrt{2}}{2}} \times (-1) = 2e^{-\frac{\sqrt{2}}{2}} \).
Numerically: \( a \approx 0.986137 \).
\( \boxed{0.986} \ \text{ms}^{-2} \).