3 attempts remaining. The equation $y=-14 \cos (5 t)$ represents the motion of a weight hanging on a spring from the ceiling after it has been stretched 14 inches and released (ignoring air resistance and friction). The outpur, $y$, gives the position of the weight in inches (in) above (positive values of $y$ ) or below (negative values of $y$ ) the equilibrium point after $t$ seconds (s). Find the first four positive times when the weight is 4.7 inches below the equilibrium. Note: make sure to put them in order! $t_{1}$ should be the first time the mass reaches equilibrium. \[ \begin{array}{l} t_{1}= \\ t_{2}= \\ t_{3}= \\ t_{4}= \\ \end{array} \] Submit answer
Solution
Step 1 ：The question is asking for the times when the weight is 4.7 inches below the equilibrium. This means we are looking for the values of \(t\) when \(y = -4.7\). So we need to solve the equation \(-14 \cos (5 t) = -4.7\) for \(t\).
Step 2 ：We can start by isolating \(\cos (5 t)\) on one side of the equation. Then we can use the inverse cosine function to solve for \(5t\). Finally, we can divide by 5 to solve for \(t\).
Step 3 ：Since the cosine function is periodic with period \(2\pi\), there will be infinitely many solutions. However, we only need the first four positive solutions.
Step 4 ：The first four positive times when the weight is 4.7 inches below the equilibrium are approximately 34.94 seconds, 38.71 seconds, 42.48 seconds, and 46.25 seconds.
Step 5 ：\[\boxed{t_{1}= 34.94, t_{2}= 38.71, t_{3}= 42.48, t_{4}= 46.25}\]
