IB Mathematics SL 5.5 integration as anti-differentiation AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme-style working
(a) First rest time \(q\)
Velocity \(v(t)=\dfrac{ds}{dt}=5.2\cdot4\cos(4t+6)=20.8\cos(4t+6)\).
At rest \(\Rightarrow v(t)=0 \Rightarrow \cos(4t+6)=0\).
General solutions: \(4t+6=\dfrac{\pi}{2}+n\pi \ \ (n\in\mathbb{Z})\).
Hence \(t=\dfrac{\tfrac{\pi}{2}-6+n\pi}{4}\). Using technology to find the first solution in \(0<t\le10\):
\(\boxed{q\approx 4.05165\ \text{s}}\) (accept \(4.05\) s to 3 s.f.). M1 A1
At rest \(\Rightarrow v(t)=0 \Rightarrow \cos(4t+6)=0\).
General solutions: \(4t+6=\dfrac{\pi}{2}+n\pi \ \ (n\in\mathbb{Z})\).
Hence \(t=\dfrac{\tfrac{\pi}{2}-6+n\pi}{4}\). Using technology to find the first solution in \(0<t\le10\):
\(\boxed{q\approx 4.05165\ \text{s}}\) (accept \(4.05\) s to 3 s.f.). M1 A1
(b) Total distance in the first \(q\) seconds
Method 1 (integral of speed, as per MS):
Speed is \(|v(t)|=|20.8\cos(4t+6)|\). Then \[ \text{Distance}=\int_{0}^{q}|v(t)|\,dt \;\; \text{(evaluate with GDC over }[0,q]\text{)} \;=\;8.51841\ldots\ \text{m}. \] So \(\boxed{8.52\text{ m}}\) (to 3 s.f.). M1 A1
Speed is \(|v(t)|=|20.8\cos(4t+6)|\). Then \[ \text{Distance}=\int_{0}^{q}|v(t)|\,dt \;\; \text{(evaluate with GDC over }[0,q]\text{)} \;=\;8.51841\ldots\ \text{m}. \] So \(\boxed{8.52\text{ m}}\) (to 3 s.f.). M1 A1
Method 2 (initial–minimum displacement difference, as per MS):
\(s(0)=5.2\sin(6)=3.31841\ldots\) (from calculator). The minimum displacement in the interval is \(-5.2\) (amplitude \(5.2\)).
Total distance \(=\;s(0)-s_{\min}=3.31841\ldots-(-5.2)=8.51841\ldots\approx \boxed{8.52\text{ m}}.\) M1 A1
\(s(0)=5.2\sin(6)=3.31841\ldots\) (from calculator). The minimum displacement in the interval is \(-5.2\) (amplitude \(5.2\)).
Total distance \(=\;s(0)-s_{\min}=3.31841\ldots-(-5.2)=8.51841\ldots\approx \boxed{8.52\text{ m}}.\) M1 A1
Total Marks: 5