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

(a)

In the model

\(h=16-15\cos\theta\),

the coefficient of \(\cos\theta\) gives the amplitude of the height function. This is equal to the radius of the Ferris wheel.

✅ Answer: \(15\text{ metres}\)

(b)(i)

The angle rotated is obtained by integrating the angular speed:

\(\theta=\displaystyle\int_0^{4.5}\dfrac{d\theta}{dt}\,dt\).

Therefore,

\(\theta=\displaystyle\int_0^{4.5}\dfrac{\pi^2}{24}\left|\sin\left(\dfrac{\pi}{2}t\right)\right|dt\).

The absolute value is needed because the expression represents the speed of rotation, which is non-negative.

Evaluating the integral gives

\(\theta=1.12387\ldots\text{ radians}\).

Equivalently,

\(\theta=\dfrac{\pi(10-\sqrt2)}{24}\).

✅ Answer: \(1.12\text{ radians}\)

(b)(ii)

Differentiate the height model with respect to \(\theta\):

\(\dfrac{dh}{d\theta}=15\sin\theta\).

Using the chain rule,

\(\dfrac{dh}{dt}=\dfrac{dh}{d\theta}\times\dfrac{d\theta}{dt}\).

When \(t=4.5\),

\(\theta=1.12387\ldots\).

Also,

\(\dfrac{d\theta}{dt}=\dfrac{\pi^2}{24}\left|\sin\left(\dfrac{\pi}{2}(4.5)\right)\right|\)

\(\dfrac{d\theta}{dt}=\dfrac{\pi^2}{24}\left|\sin\left(\dfrac{9\pi}{4}\right)\right|\)

\(\dfrac{d\theta}{dt}=0.290786\ldots\text{ radians per minute}\).

Therefore,

\(\dfrac{dh}{dt}=15\sin(1.12387\ldots)(0.290786\ldots)\)

\(\dfrac{dh}{dt}=3.93338\ldots\).

The positive value shows that Younsue is moving upwards at this time.

✅ Answer: \(\dfrac{dh}{dt}=3.93\text{ metres per minute}\)