- In a sinusoidal model, amplitude measures the vertical distance from the midline to an extreme. - Look at the coefficient multiplying the sine function. - Look at the constant added outside the sine function. - Interpret both quantities using the depth units given in the model.

1. The amplitude is the absolute value of the sine coefficient, so the amplitude is \(3\,\text{ft}\). This means the modeled depth varies up to \(3\,\text{ft}\) above or below its central level. 2. The vertical shift is \(12\), so the midline is \(H=12\,\text{ft}\). This is the model's central water depth.

Amplitude: \(3\,\text{ft}\), meaning the depth varies \(3\,\text{ft}\) above or below the central level. Midline: \(H=12\,\text{ft}\), the model's central water depth.

- The period tells how long one full oscillation takes. - One hertz means one complete cycle per second. - Ask how many cycles of the stated duration fit into one second.

1. Frequency is the reciprocal of period, so \(f=\frac{1}{0.25}=4\,\text{Hz}\). 2. A frequency of \(4\,\text{Hz}\) means the part completes \(4\) full oscillations each second.

\(4\,\text{Hz}\); the part completes \(4\) full oscillations per second.

- Identify the coefficient multiplying \(t\) inside the cosine function. - Use the standard relationship between that coefficient and the length of one complete cosine cycle.

1. For \(\cos(bt)\), the period is \(\frac{2\pi}{|b|}\). 2. Here \(b=\frac{\pi}{4}\), so the period is \(\frac{2\pi}{\pi/4}=8\,\text{s}\). 3. The model repeats the same pattern every \(8\,\text{s}\).

Period: \(8\,\text{s}\). The modeled sensor-reading pattern repeats every \(8\,\text{s}\).

- Read the greatest and least buoy heights from the vertical scale. - Find the height halfway between the two extremes. - The amplitude measures the rise above the average height, not the total range.

1. The highest point is \(8\,\text{ft}\) and the lowest is \(4\,\text{ft}\), so the midline is \(\frac{8+4}{2}=6\,\text{ft}\). 2. The amplitude is half the height difference: \(\frac{8-4}{2}=2\,\text{ft}\).

a) \(6\,\text{ft}\) b) \(2\,\text{ft}\)

- How could you test each formula against the starting measurement? - Compare the parent sine and cosine values at that time. - Check which formula agrees with the measured displacement before looking at later times.

1. At \(t=0\), sine equals \(0\), while cosine equals \(1\). 2. Only cosine gives the measured starting height of \(1\,\text{in.}\), so \(y=\cos(t)\) fits.

\(y=\cos(t)\), because it begins at its maximum value \(1\) when \(t=0\).

- Substitute the starting time into the model. - Find cosine at a zero angle. - Include the vertical center height as well as the oscillation amount.

1. At \(t=0\), \(\cos(\pi\cdot0)=\cos(0)=1\). 2. So \(H(0)=20+8\cdot1=28\,\text{ft}\).

\(28\,\text{ft}\)

- What are the maximum and minimum values of cosine? - How does the negative coefficient affect where the temperature is highest and lowest? - Include the endpoints of the \(24\)-hour interval. - Subtract the minimum temperature from the maximum temperature.

1. The cosine function ranges from \(-1\) to \(1\). Because its coefficient is \(-15\), \(T(t)\) is smallest when \(\cos\left(\frac{\pi}{12}t\right)=1\). 2. On \([0,24]\), this occurs at \(t=0\) and \(t=24\). Therefore, \(T(0)=T(24)=-15+30=15\). 3. The lowest-temperature points are \((0,15)\) and \((24,15)\). 4. The highest temperature occurs when the cosine is \(-1\), at \(t=12\): \(T(12)=15+30=45\). 5. The temperature difference is \(45-15=30\,^{\circ}\text{F}\).

The lowest temperature occurs at \((0,15)\) and \((24,15)\). The difference between the highest and lowest temperatures is \(30\,^{\circ}\text{F}\).

- The amplitude is the greatest vertical distance from the x-axis for each pure sine wave. - At \(x=\frac{\pi}{2}\), read the y-value of each of the three curves and compare the values numerically. - Common zeros are x-intercepts shared by all three graphs.

1. Graph \(f\) has amplitude \(2\), and graph \(g\) has amplitude \(1\). 2. At \(x=\frac{\pi}{2}\), the graph gives \(f\left(\frac{\pi}{2}\right)=2\), \(g\left(\frac{\pi}{2}\right)=-1\), and \(h\left(\frac{\pi}{2}\right)=1\). Therefore, \(1=2+(-1)\), so the displayed values satisfy \(h=f+g\) with both component waves contributing nonzero values. 3. All three functions are zero at integer multiples of \(\pi\). In the displayed interval, the common zeros are \(x=0\), \(x=\pi\), and \(x=2\pi\).

a) Amplitude of \(f\): \(2\); amplitude of \(g\): \(1\) b) \(h\left(\frac{\pi}{2}\right)=1=2+(-1)=f\left(\frac{\pi}{2}\right)+g\left(\frac{\pi}{2}\right)\) c) \(x=0\), \(x=\pi\), and \(x=2\pi\)

- For each float, subtract its lowest height from its highest height. - Count how many full cycles each float completes over the displayed interval. - A smaller time for one repeat means a shorter period.

1. Float \(A\) ranges from \(7\,\text{cm}\) to \(13\,\text{cm}\), a total swing of \(6\,\text{cm}\); float \(B\) ranges from \(9\,\text{cm}\) to \(11\,\text{cm}\), a swing of \(2\,\text{cm}\). 2. Float \(B\) completes two cycles in \(6\) seconds, so its period is \(3\) seconds. Float \(A\) needs \(6\) seconds for one cycle.

a) Float \(A\), \(6\,\text{cm}\) b) Float \(B\), \(3\,\text{s}\)

- Where does an unshifted cosine curve reach its first maximum? - How does replacing \(t\) by \(t-c\) affect the time of a peak? - A repeating curve has many peaks. Why is there a restriction on \(c\)?

1. With a positive cosine coefficient, the maximum occurs when the cosine input is \(0\) after the phase shift. 2. Thus the maximum at \(t=3\) gives \(c=3\) in the allowed interval. 3. The model is \(T(t)=72+6\cos\left(\frac{\pi}{4}(t-3)\right)\).

\(c=3\); \(T(t)=72+6\cos\left(\frac{\pi}{4}(t-3)\right)\)

- The required height is the middle height in the model. - What cosine value makes the extra height term disappear? - Find the smallest positive angle where cosine has that value. - Convert the angle back into a time.

1. Set the height equal to \(15\): \(15+5\cos\left(\frac{\pi}{4}t\right)=15\). 2. Hence \(\cos\left(\frac{\pi}{4}t\right)=0\). 3. The first positive zero of cosine occurs at input \(\frac{\pi}{2}\), so \(\frac{\pi}{4}t=\frac{\pi}{2}\) and \(t=2\,\text{s}\).

\(2\,\text{s}\)

- Compare the starting heights predicted by the two formulas. - Follow the graph just to the right of time zero. - Think about the sign of sine for a small positive angle and how a plus or minus sign changes the height.

1. Both models give height \(5\,\text{ft}\) at \(t=0\), so that starting height does not distinguish them. 2. Just after \(t=0\), sine is positive. The curve moves downward from \(5\,\text{ft}\), so the sine term must be subtracted. 3. The matching model is \(h(t)=5-2\sin\left(\frac{\pi}{4}t\right)\).

\(h(t)=5-2\sin\left(\frac{\pi}{4}t\right)\), because the graphed height starts at \(5\,\text{ft}\) and then falls.

- Put the stated time into the cosine model. - Recall the cosine value at a half turn. - Subtract the prediction from the measurement in the order requested; use the sign to decide whether the model predicts too much or too little.

1. Evaluate at \(t=2\): \(h(2)=10+2\cos(\pi)=10-2=8\,\text{ft}\). 2. Subtract prediction from measurement: \(9-8=1\,\text{ft}\). 3. The measured level is higher than the predicted level, so the model underestimates the level by \(1\,\text{ft}\).

a) \(8\,\text{ft}\) b) \(1\,\text{ft}\); the model underestimates the water level.

- Use the maximum and minimum levels to find the amplitude and midline. - Find the period from the time between consecutive maxima. - Connect the period with the positive coefficient of \(x\). - Locate an increasing midline crossing to determine \(c\).

1. The maximum level is \(2.5\,\text{m}\), and the minimum is \(-0.5\,\text{m}\). 2. The amplitude is \(a=\frac{2.5-(-0.5)}{2}=1.5\,\text{m}\). 3. The midline is \(d=\frac{2.5+(-0.5)}{2}=1\,\text{m}\). 4. Consecutive maxima occur at \(x=2\) and \(x=6\), so the period is \(4\) hours. 5. As \(b>0\), \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 6. The graph crosses the midline while increasing at \(x=1\), so \(c=1\). 7. The model is \(f(x)=1.5\sin\left(\frac{\pi}{2}(x-1)\right)+1\).

a) Maximum \(2.5\,\text{m}\), minimum \(-0.5\,\text{m}\), amplitude \(1.5\,\text{m}\), midline \(1\,\text{m}\), period \(4\) hours b) \(b=\frac{\pi}{2}\), \(c=1\); \(f(x)=1.5\sin\left(\frac{\pi}{2}(x-1)\right)+1\)

- Average the maximum and minimum temperatures to find \(d\). - The amplitude is the distance from the midline to an extreme temperature. - Connect the period with the coefficient of \(t\). - Find where the graph crosses the midline while increasing.

1. The maximum temperature is \(80\,^{\circ}\text{F}\) at \(t=14\), or 2:00 p.m. The minimum is \(60\,^{\circ}\text{F}\) at \(t=2\), or 2:00 a.m. 2. The midline is \(d=\frac{80+60}{2}=70\), and the amplitude is \(a=80-70=10\). 3. With a period of \(24\) hours and \(b>0\), \(b=\frac{2\pi}{24}=\frac{\pi}{12}\). 4. The graph crosses the midline while increasing at \(t=8\), so one shift is \(c=8\).

a) Maximum \(80\,^{\circ}\text{F}\) at 2:00 p.m.; minimum \(60\,^{\circ}\text{F}\) at 2:00 a.m. b) \(a=10\), \(d=70\) c) \(b=\frac{\pi}{12}\) d) \(c=8\)

- At \(t=4\), read the graph or substitute into the model. - Compare the graph with the horizontal line \(h=40\). - Find where the graph is on or above that line during the first \(12\) hours.

1. Evaluate the model at \(t=4\): \(h(4)=10\sin\left(\frac{\pi}{6}(4-2)\right)+35=10\sin\left(\frac{\pi}{3}\right)+35=5\sqrt{3}+35\approx43.66\,\text{ft}\). 2. To find when \(h(t) \geq 40\), identify where the graph meets the horizontal line \(h=40\). In the first \(12\) hours, the intersections occur at \(t=3\) and \(t=7\). 3. The graph is at or above \(40\,\text{ft}\) for \(3 \leq t \leq 7\). Therefore, the safe interval is from 3:00 a.m. through 7:00 a.m.

a) \(35+5\sqrt{3}\approx43.66\,\text{ft}\) b) From 3:00 a.m. through 7:00 a.m., inclusive

- Find the horizontal midline. - Use the distance from the midline to a maximum. - Find the horizontal length of one complete cycle. - Use the period to find the coefficient of \(x\). - Substitute \(x=1\) into your equation and then round.

1. The graph oscillates about the midline \(y=3\), so \(d=3\). 2. The maximum is \(5\), and the minimum is \(1\), so the amplitude is \(a=2\). 3. One complete cycle has length \(8\), so the period is \(8\), and \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 4. Therefore, \(f(x)=2\sin\left(\frac{\pi}{4}x\right)+3\). 5. At \(x=1\), \(f(1)=2\sin\left(\frac{\pi}{4}\right)+3=\sqrt{2}+3\approx4.41\).

a) \(f(x)=2\sin\left(\frac{\pi}{4}x\right)+3\) b) \((3+\sqrt{2})\,\text{in.}\approx4.41\,\text{in.}\)

- For each curve, identify the greatest displacement from its midline. - Use the time required for one complete cycle of each graph. - Use \(b=\frac{2\pi}{P}\), where \(P\) is the period. - Relate frequency to the length of the period.

1. Both curves oscillate about \(y=2.5\), so \(d=2.5\). 2. For \(g\), the amplitude is \(1.5\,\text{mm}\) and the period is \(4\,\text{s}\). Therefore, \(b_g=\frac{2\pi}{4}=\frac{\pi}{2}\), and \(g(t)=1.5\sin\left(\frac{\pi}{2}t\right)+2.5\). 3. For \(h\), the amplitude is \(0.5\,\text{mm}\) and the period is \(2\,\text{s}\). Therefore, \(b_h=\frac{2\pi}{2}=\pi\), and \(h(t)=0.5\sin(\pi t)+2.5\). 4. The amplitude of \(h\) is one-third the amplitude of \(g\). The period of \(h\) is half the period of \(g\), so the frequency of \(h\) is twice the frequency of \(g\).

a) \(g(t)=1.5\sin\left(\frac{\pi}{2}t\right)+2.5\) and \(h(t)=0.5\sin(\pi t)+2.5\) b) The amplitude of \(h\) is one-third the amplitude of \(g\), and the frequency of \(h\) is twice the frequency of \(g\).

- Use repeated corresponding positions on the normalized graph to determine one full cycle. - The graph also shows which point in the cycle occurs at \(t=0\). - Use the Ferris wheel diameter and highest height, not the graph's vertical scale, to determine the amplitude and midline of the height model. - Relate the period to the coefficient inside cosine, then choose the sign of the cosine coefficient from the starting position.

1. Consecutive minima on the normalized graph occur at \(t=0\) and \(t=4\), so the period is \(4\) minutes. At \(t=0\), the graph is at a minimum, so the rider starts at the lowest point. 2. The Ferris wheel radius is \(\frac{100}{2}=50\,\text{ft}\), so the amplitude of the height model is \(50\,\text{ft}\). 3. The lowest height is \(110-100=10\,\text{ft}\). The midline is \(d=\frac{110+10}{2}=60\,\text{ft}\). 4. With period \(4\), \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 5. Because the rider starts at the lowest point, the cosine coefficient is negative. Therefore, \(h(t)=-50\cos\left(\frac{\pi}{2}t\right)+60\).

a) Period: \(4\) minutes; the rider starts at the lowest point. b) \(h(t)=-50\cos\left(\frac{\pi}{2}t\right)+60\)

- Use the highest and lowest temperatures to determine the vertical features of the cycle. - Compare matching positions in consecutive cycles to determine the repeat length. - With no horizontal shift, inspect whether the graph begins at a maximum or a minimum to determine the sign of the cosine coefficient. - Extend a known maximum time by whole periods rather than extending the graph visually.

1. The graph has maximum \(80\,^{\circ}\text{F}\) and minimum \(50\,^{\circ}\text{F}\). The amplitude is \(\frac{80-50}{2}=15\,^{\circ}\text{F}\), and the midline is \(\frac{80+50}{2}=65\,^{\circ}\text{F}\). 2. The cycle runs from a minimum at \(t=0\) to the next minimum at \(t=24\), so the period is \(24\,\text{h}\). Therefore, \(b=\frac{2\pi}{24}=\frac{\pi}{12}\). 3. With no horizontal shift, \(\cos(0)=1\). Because the graph starts at the minimum, the cosine coefficient must be negative: \(a=-15\). Thus, \(T(t)=-15\cos\left(\frac{\pi}{12}t\right)+65\). 4. A maximum occurs at \(t=12\). Repeating every \(24\) hours gives later maxima at \(t=12+24k\). The first two after \(t=24\) are \(t=36\) and \(t=60\).

a) Amplitude: \(15\,^{\circ}\text{F}\); midline: \(T=65\,^{\circ}\text{F}\); period: \(24\,\text{h}\) b) \(T(t)=-15\cos\left(\frac{\pi}{12}t\right)+65\) c) \(t=36\,\text{h}\) and \(t=60\,\text{h}\)

- Measure the greatest distance from the equilibrium line for the amplitude, and compare consecutive matching peaks for the period. - Frequency tells how many cycles occur per second, while angular frequency is the coefficient of time inside the cosine function. - The graph starts at an extreme displacement rather than at equilibrium; use that to choose between a basic sine and cosine model. - For the requested displacement, set the model equal to the target value and choose the earliest positive solution.

1. The graph reaches \(6\,\text{cm}\) above and below equilibrium, so the amplitude is \(6\,\text{cm}\). Consecutive maxima occur \(2\,\text{s}\) apart, so the period is \(2\,\text{s}\). The frequency is the reciprocal of the period: \(\frac{1}{2}\,\text{Hz}\). 2. The graph begins at a positive maximum, so a cosine model with no phase shift is appropriate. With period \(2\), \(\omega=\frac{2\pi}{2}=\pi\). Thus \(d(t)=6\cos(\pi t)\). 3. Set \(6\cos(\pi t)=-3\), so \(\cos(\pi t)=-\frac{1}{2}\). The first positive angle with cosine \(-\frac{1}{2}\) is \(\frac{2\pi}{3}\). Therefore, \(\pi t=\frac{2\pi}{3}\), giving \(t=\frac{2}{3}\,\text{s}\).

a) Amplitude: \(6\,\text{cm}\); period: \(2\,\text{s}\); frequency: \(\frac{1}{2}\,\text{Hz}\) b) \(d(t)=6\cos(\pi t)\) c) \(t=\frac{2}{3}\,\text{s}\)

- Use the overall high and low clusters rather than treating one noisy measurement as an exact extreme. - The midline is halfway between the typical high and low levels. - Compare similar peaks to estimate the repeat length. - Use the direction of the data near \(t=0\) to choose the phase of a sine model.

1. The measurements have highs near \(25\) and lows near \(15\), so the estimated midline is \(\frac{25+15}{2}=20\) and the estimated amplitude is \(\frac{25-15}{2}=5\). 2. High points occur near \(t=2\) and \(t=10\), so the estimated period is \(8\) time units. 3. For period \(8\), the angular coefficient is \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 4. The data are near the midline and increasing at \(t=0\), so a reasonable estimated model is \(S(t)=5\sin\left(\frac{\pi}{4}t\right)+20\).

Estimated midline: \(20\); estimated amplitude: \(5\); estimated period: \(8\) One reasonable model is \(S(t)=5\sin\left(\frac{\pi}{4}t\right)+20\).

- A sinusoidal pattern should repeatedly move above and below a central level. - Look for recurring peaks or troughs at approximately equal horizontal spacing. - Distinguish “not enough evidence yet” from proving that a sinusoidal model is impossible.

1. Data set a provides stronger evidence for a sinusoidal model. It repeatedly rises above and falls below a stable central level, and its peaks recur at approximately regular horizontal intervals. 2. The values are centered near \(5\), so the estimated midline is \(y=5\). Peaks occur near \(t=2\) and \(t=10\), giving an estimated period of \(8\). 3. Data set b rises throughout the displayed measurements and does not show a repeated peak-trough-midline pattern. A portion of a sinusoid can rise over a limited interval, so these observations alone are insufficient to establish repeating sinusoidal behavior.

a) Data set a; it oscillates around a stable center and repeats a peak-trough pattern at roughly regular intervals. b) Estimated period: \(8\); estimated midline: \(y=5\) c) Data set b shows only sustained increase over the observed interval, with no repeated cycle, so the data do not yet establish sinusoidal periodic behavior.

- Evaluate the model at the listed integer inputs before comparing it with the observations. - Keep the specified subtraction order for every residual. - For the tolerance test, compare absolute residuals rather than signed residuals.

1. The model values at \(t=0,1,2,3,4\) are \(10\), \(13\), \(10\), \(7\), and \(10\), respectively. 2. Subtracting model value from observed value gives residuals \(0.4\), \(-0.4\), \(-0.3\), \(0.4\), and \(0.2\). 3. The absolute residuals are \(0.4\), \(0.4\), \(0.3\), \(0.4\), and \(0.2\), so the maximum absolute residual is \(0.4\). 4. Since \(0.4\le0.5\), every listed observation is within the required tolerance. The model meets the criterion.

a) Residuals for \(t=0,1,2,3,4\): \(0.4,-0.4,-0.3,0.4,0.2\) b) \(0.4\) c) Yes. The maximum absolute residual is \(0.4\), which is at most \(0.5\).

- Use the highest and lowest table values to determine the vertical features. - The table tells you how many equal time intervals make one cycle. - A positive sine model starts at a rising midline crossing; a positive cosine model can start at a maximum. - Compare \(11\) with a time in the table separated by one full period.

1. The maximum is \(20\) and the minimum is \(12\). Thus, the amplitude is \(\frac{20-12}{2}=4\), and the midline is \(\frac{20+12}{2}=16\). 2. Each two-hour step is a quarter cycle, so the period is \(P=4\cdot2=8\). Since \(b>0\), \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 3. The value at \(t=1\) is a rising midline crossing, so \(c=1\). Thus, \(R(t)=4\sin\left(\frac{\pi}{4}(t-1)\right)+16\). 4. The maximum occurs at \(t=3\), giving the equivalent positive-cosine model \(R(t)=4\cos\left(\frac{\pi}{4}(t-3)\right)+16\). 5. One period after \(t=3\) is \(t=11\), so \(R(11)=R(3)=20\).

a) Amplitude: \(4\); midline: \(R=16\); period: \(8\) b) \(R(t)=4\sin\left(\frac{\pi}{4}(t-1)\right)+16\) c) \(R(t)=4\cos\left(\frac{\pi}{4}(t-3)\right)+16\) d) \(R(11)=20\)

- The sinusoidal component is zero wherever the solid curve meets the dashed trend line. - Use two such intersections to determine the slope and y-intercept of the dashed line. - Measure the greatest vertical separation between the solid curve and the trend line to determine the amplitude of the oscillation. - Compare matching positions of the oscillation relative to the dashed line to find the period.

1. The solid curve and dashed trend line intersect at points including \((0,1)\) and \((2,2)\). The slope is \(m=\frac{2-1}{2-0}=0.5\), and the y-intercept is \(1\). Therefore, \(g(x)=0.5x+1\). 2. At \(x=1\), the solid graph has \(f(1)=3.5\), while the trend line has \(g(1)=1.5\). Their vertical difference is \(2\), so the amplitude of \(h\) is \(2\). 3. The oscillation relative to the trend repeats every \(4\) units, so the period of \(h\) is \(4\). Thus, \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 4. Therefore, \(f(x)=0.5x+1+2\sin\left(\frac{\pi}{2}x\right)\).

a) \(g(x)=0.5x+1\) b) Amplitude \(2\), period \(4\), \(b=\frac{\pi}{2}\) c) \(f(x)=0.5x+1+2\sin\left(\frac{\pi}{2}x\right)\)

- Identify the center height and wave size of each model separately. - Compare the horizontal shift with the length of one cycle. - Compare the two sine inputs. Can a repeating-pattern identity simplify the equations? - An equation with a positive sine value may have two solutions in one turn. - Convert those angle solutions back to seconds and check the interval.

1. Both models have amplitude \(2\,\text{cm}\) and period \(\frac{2\pi}{\pi/3}=6\,\text{s}\). Float \(A\) has average height \(10\,\text{cm}\); float \(B\) has average height \(13\,\text{cm}\). The peaks of \(B\) occur \(3\) seconds later than the matching peaks of \(A\). 2. A shift of \(3\) seconds is half a period, so \(\sin\left(\frac{\pi}{3}(t-3)\right)=-\sin\left(\frac{\pi}{3}t\right)\). 3. Equating heights gives \(10+2\sin\left(\frac{\pi}{3}t\right)=13-2\sin\left(\frac{\pi}{3}t\right)\), hence \(\sin\left(\frac{\pi}{3}t\right)=\frac{3}{4}\). 4. In one cycle, sine equals \(\frac{3}{4}\) at the angles \(u=\arcsin\left(\frac{3}{4}\right)\) and \(u=\pi-\arcsin\left(\frac{3}{4}\right)\). 5. Therefore, \(t=\frac{3}{\pi}\arcsin\left(\frac{3}{4}\right)\approx0.81\,\text{s}\) or \(t=3-\frac{3}{\pi}\arcsin\left(\frac{3}{4}\right)\approx2.19\,\text{s}\).

a) Average heights: \(A=10\,\text{cm}\), \(B=13\,\text{cm}\); both amplitudes \(2\,\text{cm}\) and periods \(6\,\text{s}\). Peaks of \(B\) lag by \(3\,\text{s}\). b) \(t=\frac{3}{\pi}\arcsin\left(\frac{3}{4}\right)\approx0.81\,\text{s}\) and \(t=3-\frac{3}{\pi}\arcsin\left(\frac{3}{4}\right)\approx2.19\,\text{s}\)