A bath toy is floating stationary in water such that half of the toy is above water and half of the toy is below the water. A kid pulls the toy 3.3 cm down into the water and then lets go. The toy then proceeds to bob up and down. It bobs up and down 18 times in a minute.
NOTE: Let up be positive and down be negative for this problem
(e) Using the equation x=-3.4*cos(1.88*t), Where is the toy at time t = 7.92 second?
(f) Give meaning to your answer in part (e). Is the toy mostly above the water or below the water at this time. Explain with evidence how you know thi
(e) To find the position of the toy at t = 7.92 seconds, we substitute t into the equation x=-3.4*cos(1.88*t):
x = -3.4*cos(1.88*7.92)
x = -3.4*cos(14.9056)
x ≈ -3.4*(-0.625)
x ≈ 2.125 cm
So, at t = 7.92 seconds, the toy is approximately 2.125 cm below the water.
(f) In part (e), we found that the toy is approximately 2.125 cm below the water at t = 7.92 seconds. Since the toy started with half of it above water and half below water, it is safe to say that at this time, the toy is still mostly below the water. This is evidenced by the fact that the toy completes 18 cycles of bobbing up and down in a minute, indicating that it spends more time below the water than above it.