A camera films the motion of a swing in a park - HSC - SSCE Mathematics Advanced - Question 26 - 2023 - Paper 1

To find the function $x(t)$, we start from the given differential equation: $\frac{dx}{dt} = -1.5 \sin \left( \frac{5\pi}{4} t \right).$

Integrating both sides: $x(t) = \int -1.5 \sin \left( \frac{5\pi}{4} t \right) dt.$

Using the integral of $\, \sin(kx) \, dx$, we have: $x(t) = -1.5 \cdot \left(-\frac{4}{5\pi}\right) \cos \left( \frac{5\pi}{4} t \right) + k\; (where\; k\; is\; the\; constant\; of\; integration)$

Thus: $x(t) = \frac{6}{5\pi} \cos \left( \frac{5\pi}{4} t \right) + k.$

When $t = 0$, we have $x(0) = 11.2$ m: $11.2 = \frac{6}{5\pi} \cos(0) + k \Rightarrow k = 11.2 - \frac{6}{5\pi}.$ Therefore, we have: $x(t) = \frac{6}{5\pi} \cos \left( \frac{5\pi}{4} t \right) + \left(11.2 - \frac{6}{5\pi}\right).$