91Ó°ÊÓ

Differentiating the position function y(t) with respect to time, t, gives the velocity function v(t): $$ v(t) = \frac{dy(t)}{dt} $$ Using the given equation for y(t): $$ y(t) = (3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})t + 5.0 \mathrm{~m} $$ Differentiate it to get the velocity equation: $$ v(t) = \frac{d}{dt}[(3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})t + 5.0 \mathrm{~m}] $$ The first term is a product of a constant and a sine function, while the second term is linear, and the third term is a constant. Apply the rules of differentiation for each term: $$ v(t) = (3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s}) $$

Differentiating the velocity function v(t) with respect to time, t, gives the acceleration function a(t): $$ a(t) = \frac{dv(t)}{dt} $$ Use the equation for v(t) found in Step 1: $$ v(t) = (3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s}) $$ Differentiate it to get the acceleration equation: $$ a(t) = \frac{d}{dt}\left[(3.8 \mathrm{~m})(0.46 / \mathrm{s}) \cos (0.46 t / \mathrm{s} -0.31) - (0.2 \mathrm{~m} / \mathrm{s})\right] $$ The first term is a product of a constant and a cosine function, while the second is a constant. Apply the rules of differentiation for each term: $$ a(t) = -(3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s})^2 \sin (0.46 t / \mathrm{s} -0.31) $$

To find the time points when the acceleration is zero, set a(t) equal to zero and solve for t: $$ 0 = -(3.8 \mathrm{~m})\cdot(0.46 / \mathrm{s})^2 \sin (0.46 t / \mathrm{s} -0.31) $$ We are searching for the time interval between 0 and 30 s. The sine function will be zero when its argument, \((0.46 t / \mathrm{s} - 0.31)\), is an integer multiple of \(\pi\). Thus, we can write: $$ 0.46 t / \mathrm{s} - 0.31 = k\pi $$ Solve for t: $$ t = \frac{k\pi + 0.31}{0.46 / \mathrm{s}} $$ Since we are looking for times between 0 and 30s, plug in integer values for k and check if they fall in this range. To find an approximate lower bound, plug k=0: $$ t = \frac{0\pi + 0.31}{0.46 / \mathrm{s}} \approx 0.67 \mathrm{~s} $$ To find an approximate upper bound, plug k=20: $$ t = \frac{20\pi + 0.31}{0.46 / \mathrm{s}} \approx 43.18 \mathrm{~s} $$ Since 43.18 s is outside the range, we can conclude that the valid values for k are between 0 and 20 (excluding 20). Thus, the times when the acceleration is zero are approximately: $$ t_k = \frac{k\pi + 0.31}{0.46 / \mathrm{s}} $$ for k = 0, 1, 2, ..., 19.