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

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Question 26

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A camera films the motion of a swing in a park. Let $x(t)$ be the horizontal distance, in metres, from the camera to the seat of the swing at $t$ seconds. The seat... show full transcript

Worked Solution & Example Answer:A camera films the motion of a swing in a park - HSC - SSCE Mathematics Advanced - Question 26 - 2023 - Paper 1

Step 1

Find an expression for $x(t)$

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Answer

To find the expression for x(t)x(t), we need to integrate the rate of change of xx:

x(t)=1.5sin(5π4t)dtx(t) = \int -1.5\sin\left(\frac{5\pi}{4}t\right) dt

Using the antiderivative of sine, we have:

x(t)=1.5(45πcos(5π4t))+kx(t) = -1.5\left(-\frac{4}{5\pi}\cos\left(\frac{5\pi}{4}t\right)\right) + k

This simplifies to:

x(t)=65πcos(5π4t)+kx(t) = \frac{6}{5\pi}\cos\left(\frac{5\pi}{4}t\right) + k

Now we need to find the constant kk. When t=0t = 0, x(0)=11.2x(0) = 11.2:

11.2=65πcos(0)+k11.2 = \frac{6}{5\pi}\cos(0) + k
11.2=65π+k11.2 = \frac{6}{5\pi} + k

Thus, k=11.265πk = 11.2 - \frac{6}{5\pi}

Putting it all together, we have:

x(t)=65πcos(5π4t)+11.265πx(t) = \frac{6}{5\pi}\cos\left(\frac{5\pi}{4}t\right) + 11.2 - \frac{6}{5\pi}

Step 2

How many times does the swing reach the closest point to the camera during the first 10 seconds?

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Answer

First, we determine the period of the function x(t)x(t):

The angular frequency is given by: Angular frequency=5π4\text{Angular frequency} = \frac{5\pi}{4}

The period TT is then calculated as: T=2πAngular frequency=2π5π4=85=1.6T = \frac{2\pi}{\text{Angular frequency}} = \frac{2\pi}{\frac{5\pi}{4}} = \frac{8}{5} = 1.6

In the first 10 seconds, the number of complete periods is: 101.6=6.25\frac{10}{1.6} = 6.25

Since the swing reaches its closest point 6 times in these complete cycles.

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