A boat tied up at a dock bobs up & down with the passing waves. Vertical distance between its high & low points is 1.8 m. The cycle is repeated every 4 s. At what time or times during a cycle is...
A boat tied up at a dock bobs up & down with the passing waves. Vertical distance between its high & low points is 1.8 m.
The cycle is repeated every 4 s. At what time or times during a cycle is the instantaneous rate of change of the vertical position of the boat equal to 0 and at what times is it a maximum?
- print Print
- list Cite
Expert Answers
calendarEducator since 2018
write8 answers
starTop subjects are Literature and Math
The boat moves on top of the waves, and that movement can be described as either a cosine or sine function, depending on if the boat starts at zero or at its maximum of 0.9 meters (1.8 is the total difference, so from the top to the middle, or zero, is 1.8/2 = 0.9). Let us say that the boat is at height zero at time zero and first goes upward, so we will use a sine function to describe the height h(t) at time t.
Here is the information we are given:
Amplitude (max distance from level water) = 0.9 meters
Period (one full cycle) = 4 seconds.
The general sine equation is as follows:
amplitude*sin(2*pi*time/period)
That gives us h(t) = 0.9*sin(0.5*pi*t)
The rate of change of the height, or velocity, is v(t) = 0.9*.05*pi*cos(0.5*pi*t). We get this by taking the derivative of the position function using the chain rule. v(t) = h'(t).
The velocity, v(t), will equal zero when cos(0.5*pi*t). This is when t = 1, 3, 5, and so on (any odd non-zero integer).
The velocity will be at a maximum when cos(0.5*pi*t) = 1 (since the cosine function can't exceed 1). This is when t = 0, 4, 8, and so on (4*n, where n is any integer).
One is tempted to say that when t = 2, 6, 10 (2+4*n) are also max velocities, but in fact, they are minimum velocities, as the boat is going in the downward direction, so cos(0.5*pi*t) = -1.
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
The boat moves up and down with the passing waves. I take the height of the boat to be a sinusoidal function. And the height of the boat is zero at the start of the cycle.
The distance between the high and the low points is 1.8. So the amplitude of the vertical displacement is 1.8/2 = 0.9 m.
As the cycle is repeated every 4 s, the time period is 4s.
The equation of the motion of the boat as it bobs up and down is given by f(t) = 0.9*sin ((t - 4)*0.5*pi)
The rate of change of the vertical velocity is the derivative of f(t), f'(t) = 0.9*0.5*pi*cos ((t - 4)*0.5*pi).
The instantaneous rate of change of velocity is 0 when 0.9*0.5*pi*cos ((t - 4)*0.5*pi) = 0
or at t = 1 + n*4 and at t = 3 + n*4
The instantaneous rate of change is maximum at t = 0 + 4*n and at t= 2 + n*4
Therefore the instantaneous rate of change in the vertical position is :
0 at t = 1 + n*4 and t = 3 + n*4 and maximum at t = 0 + 4*n and t= 2 + n*4
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.