3.
a) Since Jimmy is on a swing, we can model the situation using a periodic function like sine or cosine. Since Jimmy starts from one of the high points, we can use the cosine function:
[tex]f(x)=A\cos (wx)+C[/tex]
b) It takes 1.5 seconds for jimmy to go from the high point in front of the swing to the high point behind the swing, this is half of the period, therefore:
[tex]\begin{gathered} \frac{T}{2}=1.5 \\ so\colon \\ T=3 \\ where \\ \omega=2\pi f \\ and \\ f=\frac{1}{T} \\ so\colon \\ \omega=\frac{2\pi}{3} \end{gathered}[/tex]
The equation will be given by:
[tex]\begin{gathered} H(t)=A\cos (\omega t)+C \\ where \\ A=1.5 \\ C=0.5 \\ H(t)=1.5\cos (\frac{2\pi}{3}x)+0.5 \end{gathered}[/tex]
c)
For t = 0.5:
[tex]\begin{gathered} f(0.5)=1.5\cos (\frac{2\pi}{3}(0.5))+0.5 \\ f(0.5)=1.25m \end{gathered}[/tex]
