Answer:
a. Period, T = 12.57 hours. b. Latest time = 294.16 hours, 6 a.m 12 days later
Step-by-step explanation:
For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t) = 5 + 4.6sin(0.5t). (a) What is the period of the tides in hours? (b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?
a. The period, T of the tides is gotten from ω = 2π/T, where ω is the angular frequency of the wave and T its period. Comparing the sine part of equation of the tides with the general equqtion of a sine wave, Asinωt, thus Asinωt = 4.6sin(0.5t),
so ω = 0.5 rad/s.
Equating this to 2π/T implies
0.5 = 2π/T
therefore, T = 2π/0.5 = 12.566 hours ≈ 12.57 hours
b. The tide becomes too low if it is less than 2.5 m i.e d(t) < 2.5 m. So, equating d(t) as 2.5 in the tidal equation, we have d(t) = 5 + 4.6sin(0.5t).
2.5< 5+ 4.6sin(0.5t)
2.5-5 < 4.6sin(0.5t)
-2.5 < 4.6sin(0.5t)
-2.5/4.6 < sin(0.5t)
-0.543 < sin(0.5t)
sin⁻¹(-0.543)<0.5t
-32.92<0.5t. Since we cannot have negative time, we add 180 to -32.92 to give 147.08
147.08<0.5t
t>147.08/0.5=294.16 hours
So, if t is greater than 294.16 hours, the water will be shallow. That is, below 2.5 m. which is 12 days 6hours 8 min 38 s which is 6 a.m 12 days later
